Precalculus

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( f(x) = 3x - 7 \), find \( f(4) \).
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   \( 3(4) - 7 = 12 - 7 \). 5

2. If \( f(x) = x^2 - x \), find \( f(-3) \).
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   \( (-3)^2 - (-3) = 9 + 3 \). Keep the parentheses: \( -3^2 \) would give the wrong sign. 12

3. Find the domain of \( f(x) = \dfrac{1}{x + 2} \).
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   The denominator is zero when \( x = -2 \). All real numbers except \( -2 \), that is \( (-\infty, -2) \cup (-2, \infty) \)

4. Find the domain of \( f(x) = \sqrt{x + 5} \).
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   Require \( x + 5 \ge 0 \). \( [-5, \infty) \)

5. Is \( \{(1, 2), (3, 4), (1, 5)\} \) a function?
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   The input 1 is paired with both 2 and 5. No

6. For \( f(x) = x^2 + 1 \), simplify \( f(x + h) - f(x) \).
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   \( f(x + h) = (x + h)^2 + 1 = x^2 + 2xh + h^2 + 1 \). Subtract \( x^2 + 1 \): the \( x^2 \) and the 1 cancel. Check with \( x = 1, h = 1 \): \( f(2) - f(1) = 5 - 2 = 3 \), and \( 2(1)(1) + 1 = 3 \). Correct. \( 2xh + h^2 \)

7. Find the domain of \( f(x) = \dfrac{\sqrt{2x - 6}}{x - 5} \).
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   The root needs \( 2x - 6 \ge 0 \), so \( x \ge 3 \). The denominator needs \( x \ne 5 \). \( [3, 5) \cup (5, \infty) \)

8. Find the range of \( f(x) = -x^2 + 6x - 4 \).
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   The parabola opens downward, so it has a maximum at the vertex. The vertex is at \( x = \dfrac{-6}{2(-1)} = 3 \). \( f(3) = -9 + 18 - 4 = 5 \). Check at \( x = 0 \): \( f(0) = -4 \le 5 \). \( (-\infty, 5] \)

9. Find the domain of \( f(x) = \dfrac{1}{\sqrt{x^2 - 9}} \).
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   The root is in the denominator, so it must be strictly positive, not just nonnegative: \( x^2 - 9 \gt 0 \), which means \( x \lt -3 \) or \( x \gt 3 \). Both endpoints are excluded, since \( x = \pm 3 \) would put a zero in the denominator. \( (-\infty, -3) \cup (3, \infty) \)

10. A student claims \( f(a + b) = f(a) + f(b) \) for every function. Give a counterexample and say which functions it does hold for.
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    Take \( f(x) = x^2 \), \( a = 1 \), \( b = 2 \). \( f(1 + 2) = f(3) = 9 \), but \( f(1) + f(2) = 1 + 4 = 5 \). The claim fails. It holds for functions of the form \( f(x) = mx \), which are the functions that distribute over a sum: \( m(a + b) = ma + mb \). Functions with a constant, such as \( f(x) = 2x + 1 \), already fail: \( f(1 + 2) = 7 \) but \( f(1) + f(2) = 3 + 5 = 8 \). It fails for \( x^2 \); it holds only for \( f(x) = mx \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the domain of \( y = \sqrt{x} \).
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   The radicand cannot be negative. \( [0, \infty) \)

2. State the range of \( y = |x| \).
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   An absolute value is never negative, and every nonnegative number is reached. \( [0, \infty) \)

3. Name the asymptotes of \( y = \dfrac{1}{x} \).
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   The function blows up as \( x \to 0 \) and levels off as \( |x| \) grows. \( x = 0 \) (vertical) and \( y = 0 \) (horizontal)

4. Evaluate \( 2^x \) at \( x = -1 \).
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   A negative exponent means a reciprocal: \( 2^{-1} = \dfrac{1}{2} \). \( \dfrac{1}{2} \)

5. Which parent has domain and range all real numbers, is not a line, and has a flat spot at the origin?
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   \( y = x^2 \) has a restricted range and \( y = x \) is a line. The cubic is defined for all inputs, takes all outputs, and flattens at the origin. \( y = x^3 \)

6. Which of \( x^2 \), \( x^3 \), \( |x| \), \( 2^x \) are increasing on the whole real line?
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   \( x^2 \) and \( |x| \) decrease for negative \( x \), so they are out. \( x^3 \) and \( 2^x \)

7. Solve \( x^2 = 2^x \) for positive \( x \) and verify.
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   Try small integers: \( x = 2 \) gives \( 4 = 4 \), and \( x = 4 \) gives \( 16 = 16 \). Both check. No other positive integer works: \( x = 3 \) gives \( 9 \ne 8 \), and \( x = 5 \) gives \( 25 \ne 32 \). \( x = 2 \) and \( x = 4 \)

8. For \( 0 \lt x \lt 1 \), which is larger, \( \sqrt{x} \) or \( x \)? Test and explain.
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   Take \( x = 0.25 \): \( \sqrt{0.25} = 0.5 \), which exceeds 0.25. Take \( x = 0.81 \): \( \sqrt{0.81} = 0.9 \gt 0.81 \). Squaring a number between 0 and 1 makes it smaller, so taking the root makes it larger. For \( x \gt 1 \) the order reverses: \( \sqrt{4} = 2 \lt 4 \). \( \sqrt{x} \gt x \) on \( (0, 1) \)

9. Give the points of \( y = \sqrt[3]{x} \) at \( x = -8, -1, 0, 1, 8 \).
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   \( \sqrt[3]{-8} = -2 \), \( \sqrt[3]{-1} = -1 \), \( 0 \), \( 1 \), \( \sqrt[3]{8} = 2 \). \( (-8, -2), (-1, -1), (0, 0), (1, 1), (8, 2) \)

10. Why does \( \sqrt{x} \) have a restricted domain but \( \sqrt[3]{x} \) does not?
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    A square root asks for a number whose square is \( x \). Squares of real numbers are never negative, so no real number answers for a negative \( x \). A cube root asks for a number whose cube is \( x \). Cubes can be negative, since \( (-2)^3 = -8 \), so every real \( x \) has exactly one real cube root. The rule generalizes: even roots need a nonnegative radicand, odd roots do not. Even powers cannot be negative; odd powers can

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Describe the graph of \( y = f(x - 4) + 2 \) relative to \( y = f(x) \).
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   The \( x - 4 \) moves the graph to the right by 4, and the \( +2 \) moves it up by 2. Right 4, up 2

2. What does \( y = -f(x) \) do to the graph of \( f \)?
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   Every output changes sign. Reflection in the \( x \)-axis

3. What does \( y = f(-x) \) do to the graph of \( f \)?
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   Every input changes sign before the function sees it. Reflection in the \( y \)-axis

4. The point \( (3, 5) \) is on \( y = f(x) \). Find the corresponding point on \( y = f(x + 2) - 1 \).
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   The \( x + 2 \) shifts left by 2: \( x \) becomes 1. The \( -1 \) lowers \( y \) to 4. \( (1, 4) \)

5. The point \( (3, 5) \) is on \( y = f(x) \). Find the corresponding point on \( y = 3f(x) \).
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   The factor 3 multiplies the output only. \( (3, 15) \)

6. The point \( (4, 8) \) is on \( y = f(x) \). Find the corresponding point on \( y = f(2x) \).
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   We need \( 2x = 4 \), so \( x = 2 \); the output stays 8. \( (2, 8) \)

7. Write the equation of the parabola obtained from \( y = x^2 \) by shifting 3 left, reflecting in the \( x \)-axis and shifting up 5.
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   Left 3: \( (x + 3)^2 \). Reflect: \( -(x + 3)^2 \). Up 5: add 5. Vertex \( (-3, 5) \), opening downward. Check at \( x = -3 \): the value is 5. Correct. \( y = -(x + 3)^2 + 5 \)

8. The point \( (4, 1) \) is on \( y = f(x) \). Find the corresponding point on \( y = f(2x - 6) \).
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   We need the inside to equal 4: \( 2x - 6 = 4 \), so \( x = 5 \). The output stays 1. Confirm with the factored form \( f\big(2(x - 3)\big) \): halve the original \( x \), \( 4/2 = 2 \), then shift right by 3 to get 5. Both routes agree. \( (5, 1) \)

9. Find the domain and range of \( y = -3\sqrt{x - 2} + 4 \).
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   Domain: \( x - 2 \ge 0 \), so \( x \ge 2 \). The root is at least 0, so \( -3\sqrt{x - 2} \le 0 \) and the whole expression is at most 4. The largest value, 4, occurs at \( x = 2 \). Domain \( [2, \infty) \), range \( (-\infty, 4] \)

10. Does it matter whether a vertical stretch or a vertical shift is applied first? Test with \( f(x) = x \) at \( x = 1 \), comparing \( 2f(x) + 3 \) with \( 2\big(f(x) + 3\big) \).
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    \( 2f(1) + 3 = 2 + 3 = 5 \), but \( 2\big(f(1) + 3\big) = 2(4) = 8 \). They differ, so the order matters, and the notation records it. In \( a\,f(x) + k \) the stretch is applied to the function and the shift afterward. The second expression is really \( 2f(x) + 6 \), a shift of 6. The same caution applies to the inside: the order of a horizontal stretch and a shift is settled by factoring so that \( x \) stands alone. Order matters; the written form fixes it

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Is \( f(x) = x^2 \) even, odd or neither?
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   \( f(-x) = (-x)^2 = x^2 = f(x) \). Even

2. Is \( f(x) = x^5 - x \) even, odd or neither?
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   \( f(-x) = -x^5 + x = -(x^5 - x) = -f(x) \). Odd

3. Is \( f(x) = x^2 + 2x \) even, odd or neither?
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   \( f(-x) = x^2 - 2x \), which is neither \( f(x) \) nor \( -f(x) \). Neither

4. The point \( (3, 7) \) lies on the graph of an even function. Name another point.
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   An even function has \( f(-3) = f(3) = 7 \). \( (-3, 7) \)

5. The point \( (3, 7) \) lies on the graph of an odd function. Name another point.
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   An odd function has \( f(-3) = -f(3) = -7 \). \( (-3, -7) \)

6. Classify \( f(x) = |x| - x^2 \).
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   \( f(-x) = |-x| - (-x)^2 = |x| - x^2 = f(x) \). Even

7. Classify \( f(x) = x^3 - x^2 \), and explain the numerical check.
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   \( f(-x) = -x^3 - x^2 \). This is not \( f(x) \), and \( -f(x) = -x^3 + x^2 \) is not it either. Numerical check: \( f(1) = 0 \) and \( f(-1) = -1 - 1 = -2 \). For even we would need \( f(-1) = 0 \), and for odd we would need \( f(-1) = -0 = 0 \). Neither holds. Neither

8. Classify \( g(x) = \dfrac{x^3}{x^2 + 4} \).
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   \( g(-x) = \dfrac{(-x)^3}{(-x)^2 + 4} = \dfrac{-x^3}{x^2 + 4} = -g(x) \). Odd

9. Prove that the sum of two odd functions is odd.
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   Let \( f \) and \( g \) be odd and \( s(x) = f(x) + g(x) \). \( s(-x) = f(-x) + g(-x) = -f(x) - g(x) = -\big(f(x) + g(x)\big) = -s(x) \). So \( s \) is odd

10. Show that the only function that is both even and odd is the zero function, and explain what this says about an odd function at 0.
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    If \( f \) is even, \( f(-x) = f(x) \). If it is also odd, \( f(-x) = -f(x) \). Equating: \( f(x) = -f(x) \), so \( 2f(x) = 0 \) and \( f(x) = 0 \) for every \( x \). For an odd function alone, put \( x = 0 \): \( f(0) = -f(0) \), so \( f(0) = 0 \), provided 0 is in the domain. This is why \( f(x) = x + 1 \) cannot be odd: \( f(0) = 1 \). Only \( f(x) = 0 \); an odd function must pass through the origin if defined there

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( f(x) = \begin{cases} 2x & x \lt 0 \\ x + 1 & x \ge 0 \end{cases} \), find \( f(-2) \), \( f(0) \) and \( f(3) \).
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   \( f(-2) = 2(-2) = -4 \). At 0 the second rule applies: \( f(0) = 1 \). \( f(3) = 3 + 1 = 4 \). \( -4, 1, 4 \)

2. Solve \( |x| = 7 \).
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   Two numbers are at distance 7 from 0. \( x = 7 \) or \( x = -7 \)

3. Solve \( |x - 2| = 5 \).
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   \( x - 2 = 5 \) gives 7, and \( x - 2 = -5 \) gives \( -3 \). \( x = 7 \) or \( x = -3 \)

4. Solve \( |x + 1| \lt 3 \).
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   \( -3 \lt x + 1 \lt 3 \), so subtract 1 throughout. \( -4 \lt x \lt 2 \)

5. Evaluate \( |-8| - |3 - 10| \).
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   \( 8 - |-7| = 8 - 7 \). 1

6. Solve \( |3x - 1| = 11 \).
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   \( 3x - 1 = 11 \) gives \( x = 4 \). \( 3x - 1 = -11 \) gives \( x = -\dfrac{10}{3} \). Check the second: \( 3\left(-\dfrac{10}{3}\right) - 1 = -11 \), and \( |-11| = 11 \). \( x = 4 \) or \( x = -\dfrac{10}{3} \)

7. Solve \( |2x - 4| \ge 6 \).
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   \( 2x - 4 \ge 6 \) gives \( x \ge 5 \). \( 2x - 4 \le -6 \) gives \( x \le -1 \). Test \( x = 0 \): \( |-4| = 4 \), not at least 6, so 0 is correctly outside. \( x \le -1 \) or \( x \ge 5 \)

8. Solve \( |x - 3| = 2x \), and identify any extraneous solution.
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   Case 1: \( x - 3 = 2x \) gives \( x = -3 \). Case 2: \( 3 - x = 2x \) gives \( x = 1 \). Check \( x = -3 \): the left side is \( |-6| = 6 \), the right side is \( 2(-3) = -6 \). They are not equal, so \( x = -3 \) is extraneous: a distance cannot equal a negative number. Check \( x = 1 \): \( |-2| = 2 \) and \( 2(1) = 2 \). Correct. \( x = 1 \) only

9. Write \( g(x) = |x - 4| + |x + 1| \) as a piecewise function and check it at \( x = -2 \).
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   The pieces change at \( x = -1 \) and \( x = 4 \). For \( x \lt -1 \): \( (4 - x) + (-x - 1) = 3 - 2x \). For \( -1 \le x \le 4 \): \( (4 - x) + (x + 1) = 5 \). For \( x \gt 4 \): \( (x - 4) + (x + 1) = 2x - 3 \). Check at \( x = -2 \): directly \( |-6| + |-1| = 7 \), and \( 3 - 2(-2) = 7 \). Correct. \( g(x) = 3 - 2x \) for \( x \lt -1 \); \( 5 \) for \( -1 \le x \le 4 \); \( 2x - 3 \) for \( x \gt 4 \)

10. Extend the tax function of the worked example: above \$40,000 the rate is 22 percent on the excess. Find the tax on \$50,000 and explain why the function is continuous.
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    At \$40,000 the tax is \( 1000 + 0.12(30000) = 1000 + 3600 = 4600 \). For \( x \gt 40000 \): \( T(x) = 4600 + 0.22(x - 40000) \). \( T(50000) = 4600 + 0.22(10000) = 4600 + 2200 = 6800 \). Continuity: at \$40,000 the third piece gives \( 4600 + 0 \), matching the second. Each new piece starts from the total accumulated so far, which is why the function has corners and no jumps. \$6,800

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( f(x) = x + 5 \) and \( g(x) = 2x \), find \( (f \circ g)(3) \).
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   \( g(3) = 6 \), then \( f(6) = 11 \). 11

2. With the same functions, find \( (g \circ f)(3) \).
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   \( f(3) = 8 \), then \( g(8) = 16 \). Note that it differs from the previous answer. 16

3. If \( f(x) = x^2 \) and \( g(x) = x + 1 \), find \( f(g(x)) \).
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   Substitute \( x + 1 \) for \( x \) in \( x^2 \). \( (x + 1)^2 = x^2 + 2x + 1 \)

4. Find the inverse of \( f(x) = 4x - 8 \).
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   \( x = 4y - 8 \), so \( y = \dfrac{x + 8}{4} \). Check: \( f(4) = 8 \), and \( f^{-1}(8) = \dfrac{16}{4} = 4 \). \( f^{-1}(x) = \dfrac{x + 8}{4} \)

5. Is \( f(x) = x^2 \) one-to-one on all real numbers?
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   \( f(2) = 4 = f(-2) \). Two inputs share an output, so it fails the horizontal line test. No

6. Find the inverse of \( f(x) = x^3 - 5 \).
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   \( x = y^3 - 5 \), so \( y^3 = x + 5 \) and \( y = \sqrt[3]{x + 5} \). Check: \( f(3) = 22 \), and \( \sqrt[3]{27} = 3 \) at input 22. \( f^{-1}(x) = \sqrt[3]{x + 5} \)

7. If \( f(x) = \sqrt{x - 2} \) and \( g(x) = x^2 + 2 \), find \( f(g(x)) \) and its domain.
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   \( f(g(x)) = \sqrt{x^2 + 2 - 2} = \sqrt{x^2} \). This is \( |x| \), not \( x \): the square root always returns the nonnegative root. The inside \( x^2 \ge 0 \) for every real \( x \), so the domain is all reals. Check at \( x = -3 \): \( g(-3) = 11 \), \( f(11) = \sqrt{9} = 3 = |-3| \). \( |x| \), domain all real numbers

8. Find the inverse of \( f(x) = \dfrac{2x - 1}{x + 3} \).
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   \( x = \dfrac{2y - 1}{y + 3} \), so \( xy + 3x = 2y - 1 \). Collect: \( xy - 2y = -1 - 3x \), so \( y(x - 2) = -(3x + 1) \). \( y = \dfrac{-(3x + 1)}{x - 2} = \dfrac{3x + 1}{2 - x} \). Check: \( f(1) = \dfrac{1}{4} \), and \( f^{-1}\left(\tfrac14\right) = \dfrac{\frac34 + 1}{2 - \frac14} = \dfrac{7/4}{7/4} = 1 \). \( f^{-1}(x) = \dfrac{3x + 1}{2 - x} \)

9. Restrict \( f(x) = x^2 - 4x \) to \( x \ge 2 \) and find its inverse.
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   Complete the square: \( f(x) = (x - 2)^2 - 4 \). With \( x \ge 2 \), swap: \( x = (y - 2)^2 - 4 \), so \( y - 2 = \sqrt{x + 4} \) (positive, because \( y \ge 2 \)). Check: \( f(5) = 25 - 20 = 5 \), and \( 2 + \sqrt{5 + 4} = 5 \). Correct. \( f^{-1}(x) = 2 + \sqrt{x + 4} \), domain \( [-4, \infty) \)

10. Show that \( f(x) = x^2 \) and \( g(x) = \sqrt{x} \) satisfy \( f(g(x)) = x \) but are not inverses on all reals.
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    \( f(g(x)) = (\sqrt{x})^2 = x \), valid for \( x \ge 0 \), where \( g \) is defined. But \( g(f(x)) = \sqrt{x^2} = |x| \), which equals \( x \) only for \( x \ge 0 \). At \( x = -3 \): \( g(f(-3)) = \sqrt{9} = 3 \ne -3 \). Inverses need both compositions to return the input on the whole domain, and \( x^2 \) on all reals is not one-to-one. On \( [0, \infty) \) both hold. Both compositions must give \( x \); here only one does on all reals

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the average rate of change of \( f(x) = 3x + 1 \) on \( [1, 5] \).
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   \( f(1) = 4 \) and \( f(5) = 16 \), so \( \dfrac{12}{4} = 3 \). A line has the same rate everywhere. 3

2. Find the average rate of change of \( f(x) = x^2 \) on \( [1, 3] \).
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   \( \dfrac{9 - 1}{3 - 1} = \dfrac{8}{2} \). 4

3. Find the average rate of change of \( f(x) = x^2 - x \) on \( [0, 4] \).
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   \( f(0) = 0 \) and \( f(4) = 12 \), so \( \dfrac{12}{4} \). 3

4. Simplify the difference quotient for \( f(x) = 2x + 5 \).
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   \( f(x + h) - f(x) = 2(x + h) + 5 - 2x - 5 = 2h \). Divide by \( h \). 2

5. Simplify the difference quotient for \( f(x) = x^2 \).
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   \( \dfrac{(x + h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} \). \( 2x + h \)

6. Find the average rate of change of \( f(x) = \dfrac{1}{x} \) on \( [1, 4] \).
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   \( f(1) = 1 \) and \( f(4) = \dfrac{1}{4} \), so \( \dfrac{\frac14 - 1}{3} = \dfrac{-\frac34}{3} = -\dfrac{1}{4} \). Negative because the function is decreasing. \( -\dfrac{1}{4} \)

7. Simplify the difference quotient for \( f(x) = x^3 \).
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   \( (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \). Subtract \( x^3 \) and divide by \( h \). Check with \( x = 1, h = 1 \): \( \dfrac{8 - 1}{1} = 7 \), and \( 3 + 3 + 1 = 7 \). Correct. \( 3x^2 + 3xh + h^2 \)

8. For \( f(x) = \sqrt{x} \), find the average rate on \( [4, 9] \), then simplify \( \dfrac{\sqrt{4 + h} - 2}{h} \) by rationalizing.
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   On \( [4, 9] \): \( \dfrac{3 - 2}{5} = \dfrac{1}{5} \). For the quotient multiply by the conjugate: \( \dfrac{(\sqrt{4 + h} - 2)(\sqrt{4 + h} + 2)}{h(\sqrt{4 + h} + 2)} = \dfrac{(4 + h) - 4}{h(\sqrt{4 + h} + 2)} = \dfrac{1}{\sqrt{4 + h} + 2} \). As \( h \) becomes small this approaches \( \dfrac{1}{4} \). Check at \( h = 5 \): the first gives \( \dfrac{\sqrt{9} - 2}{5} = \dfrac{1}{5} \), and \( \dfrac{1}{3 + 2} = \dfrac15 \). \( \dfrac{1}{5} \); the quotient is \( \dfrac{1}{\sqrt{4 + h} + 2} \)

9. Simplify the difference quotient for \( f(x) = \dfrac{1}{x} \).
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   \( \dfrac{\frac{1}{x + h} - \frac{1}{x}}{h} \). Common denominator in the numerator: \( \dfrac{x - (x + h)}{x(x + h)} = \dfrac{-h}{x(x + h)} \). Divide by \( h \): \( -\dfrac{1}{x(x + h)} \). Check with \( x = 1, h = 1 \): \( \dfrac{\frac12 - 1}{1} = -\dfrac12 \), and \( -\dfrac{1}{1 \cdot 2} = -\dfrac12 \). Correct. \( -\dfrac{1}{x(x + h)} \)

10. For \( s(t) = -16t^2 + 64t \), the average velocity on \( [1, 3] \) is 0. Explain what this means, and find the average velocity on \( [1, 1.01] \) and the apparent instantaneous velocity at \( t = 1 \).
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    An average of zero means the ball ends at the height where it began over that interval. It does not mean the ball was at rest: it rose and fell. \( s(1) = 48 \) and \( s(1.01) = -16(1.0201) + 64.64 = 48.3184 \). Average velocity: \( \dfrac{0.3184}{0.01} = 31.84 \) ft/s. The exact difference quotient is \( -32t - 16h + 64 \), which at \( t = 1 \) is \( 32 - 16h \); for \( h = 0.01 \) that is 31.84, matching, and as \( h \to 0 \) it approaches 32. Zero average, but 31.84 ft/s on the short interval and 32 ft/s at the instant

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Describe the end behavior of \( y = 3x^5 - x \).
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   Odd degree, positive leading coefficient. Down on the left, up on the right

2. Describe the end behavior of \( y = -2x^4 + x \).
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   Even degree, negative leading coefficient. Down on both ends

3. Find the zeros of \( (x - 2)(x + 5) \).
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   Set each factor to zero. \( x = 2 \) and \( x = -5 \)

4. State the multiplicity of the zero \( x = 4 \) in \( (x - 4)^3(x + 1) \), and whether the graph crosses.
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   The exponent is 3, which is odd. Multiplicity 3; the graph crosses, flattening as it does

5. Find the degree of \( (x - 1)^2(x + 3)(x - 7)^3 \).
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   Add the exponents: \( 2 + 1 + 3 \). 6

6. Find the zeros of \( f(x) = x^3 - 4x^2 + 4x \) with their multiplicities.
   Show the full solution

   Factor out \( x \): \( x(x^2 - 4x + 4) = x(x - 2)^2 \). Check at \( x = 1 \): the original gives \( 1 - 4 + 4 = 1 \), and \( 1(1 - 2)^2 = 1 \). \( x = 0 \) (multiplicity 1) and \( x = 2 \) (multiplicity 2)

7. Write a cubic with zeros \( -1, 2, 4 \) that passes through \( (0, 16) \).
   Show the full solution

   \( P(x) = a(x + 1)(x - 2)(x - 4) \). Then \( P(0) = a(1)(-2)(-4) = 8a = 16 \), so \( a = 2 \). Check \( P(3) = 2(4)(1)(-1) = -8 \) and the sign pattern: the polynomial is positive after 4, negative between 2 and 4, positive between \( -1 \) and 2. At 3 it is negative. \( P(x) = 2(x + 1)(x - 2)(x - 4) \)

8. What is the largest number of turning points a degree 5 polynomial can have? Give a degree 3 polynomial with none.
   Show the full solution

   At most \( n - 1 = 4 \). The bound is not always reached: \( y = x^3 \) has degree 3 and no turning points, since it rises throughout and only flattens at the origin. 4; \( y = x^3 \)

9. Find the sign of \( P(x) = (x + 3)(x - 1)^2(x - 4) \) on each interval.
   Show the full solution

   Zeros at \( -3, 1, 4 \). Test one point in each interval: \( P(-4) = (-1)(25)(-8) = 200 \), positive. \( P(0) = (3)(1)(-4) = -12 \), negative. \( P(2) = (5)(1)(-2) = -10 \), negative. \( P(5) = (8)(16)(1) = 128 \), positive. The sign is the same on both sides of \( x = 1 \), because 1 has even multiplicity. Positive on \( (-\infty, -3) \) and \( (4, \infty) \); negative on \( (-3, 1) \) and \( (1, 4) \)

10. Explain why every odd-degree polynomial has at least one real zero, and why an even-degree one need not.
    Show the full solution

    An odd-degree polynomial has ends that go in opposite directions, one toward \( +\infty \) and one toward \( -\infty \). It is continuous, so it cannot jump from negative to positive, and must pass through zero somewhere. An even-degree polynomial has both ends on the same side. \( y = x^2 + 1 \) has both ends up and its minimum is 1, so it never reaches the axis. Opposite ends force a crossing; same-side ends do not

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the remainder when \( x^2 + 3x + 1 \) is divided by \( x - 2 \).
   Show the full solution

   By the remainder theorem, \( P(2) = 4 + 6 + 1 \). 11

2. Is \( x - 1 \) a factor of \( x^3 - 1 \)?
   Show the full solution

   \( P(1) = 1 - 1 = 0 \). Yes

3. Divide \( x^2 + 5x + 6 \) by \( x + 2 \).
   Show the full solution

   Factor the top: \( (x + 2)(x + 3) \). \( x + 3 \)

4. Divide \( x^3 - 6x^2 + 11x - 6 \) by \( x - 1 \) using synthetic division.
   Show the full solution

   With \( c = 1 \): bring down 1; \( -6 + 1 = -5 \); \( 11 - 5 = 6 \); \( -6 + 6 = 0 \). \( x^2 - 5x + 6 \), remainder 0

5. Use the remainder theorem to find \( P(-1) \) for \( P(x) = 2x^3 - x + 4 \).
   Show the full solution

   \( 2(-1)^3 - (-1) + 4 = -2 + 1 + 4 \). 3

6. Divide \( 3x^3 - 2x + 7 \) by \( x + 2 \).
   Show the full solution

   Insert the missing \( x^2 \) term: coefficients \( 3, 0, -2, 7 \), and \( c = -2 \). Bring down 3; \( 3 \times -2 = -6 \), \( 0 - 6 = -6 \); \( -6 \times -2 = 12 \), \( -2 + 12 = 10 \); \( 10 \times -2 = -20 \), \( 7 - 20 = -13 \). Check: \( P(-2) = -24 + 4 + 7 = -13 \). Correct. \( 3x^2 - 6x + 10 \), remainder \( -13 \)

7. Find \( k \) so that \( x - 3 \) is a factor of \( x^3 - 4x^2 + kx + 12 \).
   Show the full solution

   \( P(3) = 27 - 36 + 3k + 12 = 3 + 3k = 0 \), so \( k = -1 \). Check: \( 27 - 36 - 3 + 12 = 0 \). \( k = -1 \)

8. Factor \( x^3 - 7x + 6 \) completely, given that \( x = 1 \) is a zero.
   Show the full solution

   Coefficients \( 1, 0, -7, 6 \), \( c = 1 \): bring down 1; \( 0 + 1 = 1 \); \( -7 + 1 = -6 \); \( 6 - 6 = 0 \). The quotient \( x^2 + x - 6 = (x + 3)(x - 2) \). Check by expanding \( (x - 1)(x - 2)(x + 3) \): \( (x^2 - 3x + 2)(x + 3) = x^3 - 7x + 6 \). \( (x - 1)(x - 2)(x + 3) \)

9. Divide \( x^4 + 2x^3 - x + 5 \) by \( x^2 + 1 \) and state the remainder.
   Show the full solution

   Inserting \( 0x^2 \): the quotient is \( x^2 + 2x - 1 \) and the remainder is \( -3x + 6 \), as worked in the example. Check at \( x = 1 \): the left side is \( 1 + 2 - 1 + 5 = 7 \); the right side is \( (2)(2) + 3 = 7 \). Quotient \( x^2 + 2x - 1 \), remainder \( -3x + 6 \)

10. Explain why the remainder theorem is true, then use synthetic division to find \( P(5) \) for \( P(x) = x^3 - 6x^2 + 2x - 8 \) and confirm it directly.
    Show the full solution

    Division gives \( P(x) = (x - c)Q(x) + R \), where \( R \) is a constant because the divisor has degree 1. Putting \( x = c \) makes the first term zero, so \( P(c) = R \). Synthetic with \( c = 5 \): coefficients \( 1, -6, 2, -8 \); bring down 1; \( -6 + 5 = -1 \); \( 2 - 5 = -3 \); \( -8 - 15 = -23 \). Direct: \( 125 - 150 + 10 - 8 = -23 \). Correct. \( P(5) = -23 \); synthetic division is a fast way to evaluate

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. List the possible rational zeros of \( x^3 + 2x^2 - 5x - 6 \).
   Show the full solution

   The constant is \( -6 \), the leading coefficient is 1. \( \pm 1, \pm 2, \pm 3, \pm 6 \)

2. Is \( -1 \) a zero of \( x^3 + 2x^2 - 5x - 6 \)?
   Show the full solution

   \( -1 + 2 + 5 - 6 = 0 \). Yes

3. Factor \( x^3 + 2x^2 - 5x - 6 \) completely.
   Show the full solution

   Divide by \( x + 1 \) (using \( c = -1 \)): coefficients \( 1, 2, -5, -6 \); bring down 1; \( 2 - 1 = 1 \); \( -5 - 1 = -6 \); \( -6 + 6 = 0 \). Quotient \( x^2 + x - 6 \). That factors as \( (x + 3)(x - 2) \). Check by expanding \( (x+1)(x+3)(x-2) \) at \( x = 1 \): \( 2 \cdot 4 \cdot -1 = -8 \), and the original gives \( 1 + 2 - 5 - 6 = -8 \). \( (x + 1)(x + 3)(x - 2) \)

4. Find all zeros of \( x^2 + 9 \).
   Show the full solution

   \( x^2 = -9 \). \( x = \pm 3i \)

5. A polynomial with real coefficients has the zero \( 2 - i \). Name another zero.
   Show the full solution

   Non-real zeros come in conjugate pairs. \( 2 + i \)

6. Find all zeros of \( 2x^3 + x^2 - 13x + 6 \).
   Show the full solution

   Test \( x = 2 \): \( 16 + 4 - 26 + 6 = 0 \). Synthetic with \( c = 2 \), coefficients \( 2, 1, -13, 6 \): bring down 2; \( 1 + 4 = 5 \); \( -13 + 10 = -3 \); \( 6 - 6 = 0 \). The quotient \( 2x^2 + 5x - 3 = (2x - 1)(x + 3) \). \( x = 2, \ \dfrac12, \ -3 \)

7. Find all zeros of \( x^4 - 16 \).
   Show the full solution

   Difference of squares twice: \( (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4) \). The last factor gives \( x = \pm 2i \). Four zeros for a quartic. \( x = 2, -2, 2i, -2i \)

8. Write the real polynomial of least degree with zeros 3 and \( 1 + 2i \).
   Show the full solution

   Include the conjugate \( 1 - 2i \). The pair gives \( x^2 - 2x + 5 \). Multiply by \( x - 3 \). Check at \( x = 3 \): the product is zero. Check at \( x = 0 \): \( (-3)(5) = -15 \). \( x^3 - 5x^2 + 11x - 15 \)

9. Find all zeros of \( x^4 - 2x^3 + 2x^2 - 2x + 1 \), with multiplicities.
   Show the full solution

   Test \( x = 1 \): \( 1 - 2 + 2 - 2 + 1 = 0 \). Synthetic, coefficients \( 1, -2, 2, -2, 1 \): \( 1, -1, 1, -1, 0 \). Again with \( c = 1 \) on \( 1, -1, 1, -1 \): \( 1, 0, 1, 0 \). The quotient is \( x^2 + 1 \). So the polynomial is \( (x - 1)^2(x^2 + 1) \). \( x = 1 \) (multiplicity 2), \( x = i \), \( x = -i \)

10. Explain why a cubic with real coefficients cannot have exactly one non-real zero.
    Show the full solution

    Non-real zeros of a real polynomial come in conjugate pairs, so their number is even: 0, 2, 4 and so on. A cubic has exactly 3 zeros counting multiplicity, so the number of non-real ones can only be 0 or 2, never 1. It follows that every real cubic has at least one real zero, agreeing with the end behavior argument of lesson 2.1. Non-real zeros pair up, so their count is even

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Describe the end behavior of \( y = x^3 - x^2 - 6x \).
   Show the full solution

   Odd degree, positive leading coefficient. Down on the left, up on the right

2. Find the \( y \)-intercept of \( y = (x - 2)(x + 3)(x - 5) \).
   Show the full solution

   \( (-2)(3)(-5) = 30 \). 30

3. Does the graph of \( y = (x - 1)^2(x + 2) \) cross or touch the axis at \( x = 1 \)?
   Show the full solution

   The factor \( (x - 1) \) is squared, so the multiplicity is even. It touches

4. Find the zeros of \( y = x^3 - 9x \).
   Show the full solution

   \( x(x^2 - 9) = x(x - 3)(x + 3) \). \( x = 0, 3, -3 \)

5. What is the most turning points a degree 4 polynomial can have?
   Show the full solution

   \( n - 1 \) with \( n = 4 \). 3

6. Give the end behavior, zeros, \( y \)-intercept and signs for \( y = -(x + 1)(x - 2)^2 \).
   Show the full solution

   Degree 3, leading coefficient \( -1 \): up on the left, down on the right. Zeros: \( -1 \) (crosses) and 2 (touches). \( y \)-intercept: \( -(1)(4) = -4 \). \( y(-2) = -(-1)(16) = 16 \), positive. \( y(0) = -4 \), negative. \( y(3) = -(4)(1) = -4 \), negative. The sign stays negative on both sides of 2, as even multiplicity predicts. Positive for \( x \lt -1 \); negative elsewhere except the zeros

7. Find the cubic with zeros \( -2, 1, 4 \) and \( y \)-intercept \( -16 \).
   Show the full solution

   \( P(x) = a(x + 2)(x - 1)(x - 4) \). \( P(0) = a(2)(-1)(-4) = 8a = -16 \), so \( a = -2 \). Check that the ends agree: negative leading coefficient, degree 3, so up on the left. At \( x = -3 \): \( -2(-1)(-4)(-7) = 56 \), positive. Correct. \( P(x) = -2(x + 2)(x - 1)(x - 4) \)

8. Describe \( y = x^4 - 5x^2 + 4 \): zeros, symmetry, \( y \)-intercept, end behavior.
   Show the full solution

   \( (x^2 - 1)(x^2 - 4) \), so zeros \( \pm 1, \pm 2 \). Only even powers, so the graph is symmetric about the \( y \)-axis. \( y(0) = 4 \). Up at both ends. Zeros \( \pm 1, \pm 2 \); even; \( y \)-intercept 4; up at both ends

9. Where is \( P(x) = x^3 - x^2 - 6x \) negative?
   Show the full solution

   From the sign table: negative for \( x \lt -2 \) and for \( 0 \lt x \lt 3 \). Check: \( P(-3) = -18 \) and \( P(1) = -6 \), both negative. \( (-\infty, -2) \cup (0, 3) \)

10. A student sketches a graph with zeros at \( -1, 2, 5 \), each crossing, and both ends pointing down. Explain why this cannot be a polynomial graph.
    Show the full solution

    Each crossing changes the sign, and there are three, an odd number of changes. A graph that starts below the axis and changes sign three times ends above it. Both ends down means it starts and ends on the same side, which needs an even number of sign changes. The two facts contradict. A fix: make one zero a touch (even multiplicity), or add a fourth crossing. Three sign changes force opposite ends

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the domain of \( f(x) = \dfrac{1}{x - 3} \).
   Show the full solution

   The denominator is zero when \( x = 3 \). All reals except 3

2. Find the vertical asymptote of \( f(x) = \dfrac{x + 1}{x - 4} \).
   Show the full solution

   At \( x = 4 \) the denominator is zero and the numerator is 5. \( x = 4 \)

3. Find the horizontal asymptote of \( f(x) = \dfrac{2x + 1}{x - 5} \).
   Show the full solution

   Equal degrees: the ratio of leading coefficients is \( \dfrac{2}{1} \). \( y = 2 \)

4. Find the horizontal asymptote of \( f(x) = \dfrac{x + 3}{x^2 + 1} \).
   Show the full solution

   The degree of the top (1) is less than the bottom (2). \( y = 0 \)

5. Does \( f(x) = \dfrac{x - 3}{x^2 - 9} \) have a hole or a vertical asymptote at \( x = 3 \)?
   Show the full solution

   Factor: \( \dfrac{x - 3}{(x - 3)(x + 3)} \). The factor \( x - 3 \) cancels. A hole at \( x = 3 \) (and an asymptote at \( x = -3 \))

6. Find all asymptotes of \( f(x) = \dfrac{4x^2}{x^2 - 9} \).
   Show the full solution

   \( x^2 - 9 = (x - 3)(x + 3) \) is zero at \( \pm 3 \), and the numerator is 36 there. Equal degrees with leading coefficients 4 and 1. Vertical: \( x = 3 \) and \( x = -3 \). Horizontal: \( y = 4 \)

7. Find the slant asymptote of \( f(x) = \dfrac{x^2 + 3x + 5}{x + 1} \).
   Show the full solution

   Synthetic division with \( c = -1 \), coefficients \( 1, 3, 5 \): bring down 1; \( 3 - 1 = 2 \); \( 5 - 2 = 3 \). The quotient is \( x + 2 \), remainder 3. Check at \( x = 9 \): \( \dfrac{81 + 27 + 5}{10} = 11.3 \), and \( 9 + 2 + 0.3 = 11.3 \). \( y = x + 2 \)

8. Find the coordinates of the hole in \( f(x) = \dfrac{x^2 - x - 6}{x - 3} \).
   Show the full solution

   \( \dfrac{(x - 3)(x + 2)}{x - 3} \), so the hole is at \( x = 3 \), with height \( 3 + 2 = 5 \). \( (3, 5) \)

9. Describe the behavior of \( f(x) = \dfrac{x + 2}{x - 1} \) as \( x \) approaches 1 from each side.
   Show the full solution

   The numerator is near 3. For \( x = 1.01 \): \( \dfrac{3.01}{0.01} = 301 \), large and positive. For \( x = 0.99 \): \( \dfrac{2.99}{-0.01} = -299 \), large and negative. \( +\infty \) from the right and \( -\infty \) from the left

10. Can a rational function's graph cross its horizontal asymptote? Can it cross a vertical one? Give an example for the first.
    Show the full solution

    \( f(x) = \dfrac{x}{x^2 + 1} \) has the horizontal asymptote \( y = 0 \), yet \( f(0) = 0 \) and the graph crosses it at the origin. A horizontal asymptote describes the ends only, for large \( |x| \). A vertical asymptote sits where the function is not defined, so there is no point of the graph on it to cross. Yes for horizontal (e.g. \( \frac{x}{x^2+1} \)); never for vertical

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the vertical and horizontal asymptotes of \( f(x) = \dfrac{1}{x - 2} \).
   Show the full solution

   The denominator is zero at 2; the degree of the top is lower. \( x = 2 \) and \( y = 0 \)

2. Find the \( x \)- and \( y \)-intercepts of \( f(x) = \dfrac{x - 4}{x + 1} \).
   Show the full solution

   \( x \)-intercept: \( x - 4 = 0 \). \( y \)-intercept: \( f(0) = \dfrac{-4}{1} \). \( (4, 0) \) and \( (0, -4) \)

3. Find the horizontal asymptote of \( f(x) = \dfrac{3x^2}{x^2 + 1} \).
   Show the full solution

   Equal degrees: \( \dfrac{3}{1} \). \( y = 3 \)

4. Is \( f(x) = \dfrac{x - 1}{x + 2} \) positive or negative at \( x = 0 \)?
   Show the full solution

   \( \dfrac{-1}{2} \). Negative

5. Find the \( y \)-intercept of \( f(x) = \dfrac{x + 3}{(x - 1)(x + 2)} \).
   Show the full solution

   \( f(0) = \dfrac{3}{(-1)(2)} \). \( -\dfrac32 \)

6. For \( f(x) = \dfrac{x - 1}{(x + 2)(x - 3)} \), where is \( f \) positive?
   Show the full solution

   From the sign table: positive on \( (-2, 1) \) and on \( (3, \infty) \). Check: \( f(0) = \tfrac16 \gt 0 \) and \( f(4) = \tfrac12 \gt 0 \). \( (-2, 1) \cup (3, \infty) \)

7. Write a rational function with a vertical asymptote at \( x = 1 \), horizontal asymptote \( y = 2 \) and an \( x \)-intercept at 0.
   Show the full solution

   Numerator \( x \) for the intercept, denominator \( x - 1 \) for the asymptote, and a factor of 2 for the horizontal asymptote. Check: \( f(0) = 0 \); \( f(100) = \dfrac{200}{99} \approx 2.02 \). \( f(x) = \dfrac{2x}{x - 1} \)

8. For \( g(x) = \dfrac{x^2 - 1}{x^2 - 4} \), find where \( g \) is negative.
   Show the full solution

   Boundaries \( -2, -1, 1, 2 \). \( g(-1.5) = \dfrac{1.25}{-1.75} \lt 0 \) and \( g(1.5) \lt 0 \); \( g(0) = \tfrac14 \gt 0 \); \( g(\pm 3) = \tfrac85 \gt 0 \). \( (-2, -1) \cup (1, 2) \)

9. Write a rational function with vertical asymptote \( x = 2 \), horizontal asymptote \( y = 3 \) and \( x \)-intercept \( -1 \), and find its \( y \)-intercept.
   Show the full solution

   \( f(x) = \dfrac{3(x + 1)}{x - 2} \), as built in the example. \( f(0) = \dfrac{3}{-2} \). \( f(x) = \dfrac{3(x+1)}{x-2} \), \( y \)-intercept \( -\dfrac32 \)

10. Analyze \( f(x) = \dfrac{x^2 - x - 2}{x^2 - 4} \). Explain why \( x = 2 \) is not an asymptote.
    Show the full solution

    Factor: \( \dfrac{(x - 2)(x + 1)}{(x - 2)(x + 2)} \). The factor \( x - 2 \) cancels, so \( x = 2 \) is a hole at height \( \dfrac{3}{4} \). The remaining denominator \( x + 2 \) gives a vertical asymptote at \( x = -2 \). Horizontal asymptote \( y = 1 \). \( x \)-intercept \( -1 \). \( y \)-intercept \( \dfrac{-2}{-4} = \dfrac12 \). Near \( x = 2 \) the values approach \( \dfrac34 \) from both sides: \( f(2.01) = 0.7537 \) and \( f(1.99) = 0.7463 \). They do not blow up, so it is a hole. Hole at \( (2, \tfrac34) \); asymptote at \( x = -2 \); \( y = 1 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( x^2 - 9 \lt 0 \).
   Show the full solution

   \( (x - 3)(x + 3) \lt 0 \): negative between the roots. Test \( x = 0 \): \( -9 \lt 0 \). \( -3 \lt x \lt 3 \)

2. Solve \( x^2 - 4x + 3 \ge 0 \).
   Show the full solution

   \( (x - 1)(x - 3) \ge 0 \): nonnegative outside the roots, including them. \( x \le 1 \) or \( x \ge 3 \)

3. Solve \( (x - 2)(x + 5) \gt 0 \).
   Show the full solution

   Test \( x = 0 \): \( -10 \lt 0 \), so the middle is excluded. \( x \lt -5 \) or \( x \gt 2 \)

4. Solve \( \dfrac{x + 1}{x - 3} \gt 0 \).
   Show the full solution

   Critical values \( -1 \) and 3. \( x = 0 \): \( \dfrac{1}{-3} \lt 0 \). The outer intervals are positive. \( x \lt -1 \) or \( x \gt 3 \)

5. Solve \( x^3 - x \ge 0 \).
   Show the full solution

   \( x(x - 1)(x + 1) \ge 0 \). Test: \( x = 2 \): \( 6 \gt 0 \); \( x = 0.5 \): negative; \( x = -0.5 \): \( 0.375 \gt 0 \); \( x = -2 \): negative. \( [-1, 0] \cup [1, \infty) \)

6. Solve \( \dfrac{x - 4}{x + 2} \le 0 \).
   Show the full solution

   Critical values \( -2 \) (excluded) and 4 (included). Test \( x = 0 \): \( -2 \lt 0 \). \( -2 \lt x \le 4 \)

7. Solve \( \dfrac{x}{x - 1} \ge 2 \).
   Show the full solution

   \( \dfrac{x}{x - 1} - 2 = \dfrac{x - 2x + 2}{x - 1} = \dfrac{2 - x}{x - 1} \ge 0 \), which is \( \dfrac{x - 2}{x - 1} \le 0 \). Critical values 1 (excluded) and 2 (included). Verify: \( x = 1.5 \): \( \dfrac{1.5}{0.5} = 3 \ge 2 \) ✓; \( x = 3 \): \( 1.5 \lt 2 \), excluded ✓; \( x = 2 \): \( 2 \ge 2 \) ✓. \( 1 \lt x \le 2 \)

8. Solve \( (x - 1)^2(x + 3) \ge 0 \).
   Show the full solution

   The square is never negative and is zero at 1. The sign is therefore decided by \( x + 3 \), except at \( x = 1 \) where the whole expression is 0 and the inequality holds. \( x \ge -3 \)

9. Solve \( 2x^2 + x - 6 \lt 0 \).
   Show the full solution

   Factor: \( (2x - 3)(x + 2) \lt 0 \). Roots \( \dfrac32 \) and \( -2 \). The parabola opens up, so it is negative between the roots. Check \( x = 0 \): \( -6 \lt 0 \) ✓; \( x = 2 \): \( 8 + 2 - 6 = 4 \gt 0 \), excluded ✓. \( -2 \lt x \lt \dfrac32 \)

10. A student solves \( \dfrac{3}{x - 2} \le 1 \) by multiplying by \( x - 2 \) to get \( 3 \le x - 2 \), so \( x \ge 5 \). Find the error and the correct solution.
    Show the full solution

    Multiplying by \( x - 2 \) is only valid when it is positive. When \( x \lt 2 \) it is negative and the inequality must reverse. Correct method: \( \dfrac{3}{x - 2} - 1 = \dfrac{5 - x}{x - 2} \le 0 \), which is \( \dfrac{x - 5}{x - 2} \ge 0 \). Critical values 2 (excluded) and 5 (included). Test: \( x = 0 \): \( \dfrac{-5}{-2} \gt 0 \); \( x = 3 \): \( \dfrac{-2}{1} \lt 0 \); \( x = 6 \): positive. Verify \( x = 0 \) in the original: \( \dfrac{3}{-2} = -1.5 \le 1 \) ✓. The student's answer missed all of \( x \lt 2 \). \( x \lt 2 \) or \( x \ge 5 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( f(x) = 2^x \), find \( f(5) \).
   Show the full solution

   \( 2^5 = 32 \). 32

2. Evaluate \( 3^{-2} \).
   Show the full solution

   A negative exponent means a reciprocal: \( \dfrac{1}{3^2} \). \( \dfrac19 \)

3. Is \( y = 5(0.9)^x \) growth or decay?
   Show the full solution

   The base is 0.9, which is less than 1. Decay

4. Find the \( y \)-intercept of \( y = 7 \cdot 3^x \).
   Show the full solution

   At \( x = 0 \), \( 3^0 = 1 \). \( (0, 7) \)

5. Find the horizontal asymptote of \( y = 2^x + 4 \).
   Show the full solution

   \( 2^x \) approaches 0 as \( x \to -\infty \), so the function approaches 4. \( y = 4 \)

6. A population of 500 grows 4 percent a year. Write the model and find the population after 10 years.
   Show the full solution

   \( P(t) = 500(1.04)^t \). \( P(10) = 500(1.48024) = 740.12 \). About 740

7. Find the value of \$2,000 after 8 years at 5 percent compounded quarterly.
   Show the full solution

   \( A = 2000\left(1 + \dfrac{0.05}{4}\right)^{32} = 2000(1.0125)^{32} \). \( (1.0125)^{32} = 1.48886 \), so \( A = 2977.72 \). Compare simple growth: \( 2000(1 + 0.05 \cdot 8) = 2800 \). Compounding earns more. \$2,977.72

8. Find the value of \$1,000 after 10 years at 5 percent compounded continuously.
   Show the full solution

   \( A = 1000e^{0.5} = 1000(1.64872) \). \$1,648.72

9. Give the domain, range, asymptote and one point of \( y = 2^{x - 1} + 3 \).
   Show the full solution

   Shifted right 1 and up 3 from \( 2^x \). Domain all reals. The range is \( y \gt 3 \). The asymptote is \( y = 3 \). The point \( (0, 1) \) on \( 2^x \) moves to \( (1, 4) \); check: \( 2^0 + 3 = 4 \). Domain \( \mathbb{R} \), range \( (3, \infty) \), asymptote \( y = 3 \), point \( (1, 4) \)

10. A table gives \( y = 3, 6, 12, 24 \) at \( x = 0, 1, 2, 3 \); another gives \( 3, 6, 11, 18 \). Decide which is exponential and explain.
    Show the full solution

    First table: ratios \( 6/3, 12/6, 24/12 \) are all 2, constant, so it is exponential: \( y = 3 \cdot 2^x \). Second: ratios \( 2, 1.83, 1.64 \) change, so it is not. Differences are 3, 5, 7, which grow by a constant 2, making it quadratic. The first; constant ratios mean exponential, constant differences mean linear

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( \log_2 8 \).
   Show the full solution

   \( 2^3 = 8 \). 3

2. Evaluate \( \log 1000 \).
   Show the full solution

   Base 10: \( 10^3 = 1000 \). 3

3. Evaluate \( \ln e \).
   Show the full solution

   \( e^1 = e \). 1

4. Write \( 10^3 = 1000 \) in logarithmic form.
   Show the full solution

   The base is 10, the exponent is 3. \( \log_{10} 1000 = 3 \)

5. Evaluate \( \log_7 1 \).
   Show the full solution

   \( 7^0 = 1 \). 0

6. Evaluate \( \log_9 3 \).
   Show the full solution

   \( 9^{1/2} = 3 \). \( \dfrac12 \)

7. Solve \( \log_x 64 = 3 \).
   Show the full solution

   \( x^3 = 64 \), so \( x = 4 \). Check: \( 4^3 = 64 \). \( x = 4 \)

8. Find the domain of \( y = \ln(5 - x) \).
   Show the full solution

   Require \( 5 - x \gt 0 \), so \( x \lt 5 \). Check \( x = 4 \): \( \ln 1 = 0 \) is defined; \( x = 6 \): \( \ln(-1) \) is not. \( (-\infty, 5) \)

9. Evaluate \( 5^{\log_5 12} \) and \( \log_3 3^{x + 1} \).
   Show the full solution

   Each pair is an exponential and a logarithm of the same base, which undo each other. 12 and \( x + 1 \)

10. Explain why \( \log_b(-4) \) is undefined, and why the base cannot be 1.
    Show the full solution

    \( \log_b(-4) = y \) would mean \( b^y = -4 \). A positive base raised to any real power is positive, so nothing satisfies it. For base 1, \( 1^y = 1 \) for every \( y \), so \( \log_1 x \) would be undefined for \( x \ne 1 \) and every number for \( x = 1 \): not a function. The restrictions \( b \gt 0 \) and \( b \ne 1 \) are exactly what makes \( b^x \) one-to-one and gives it an inverse. A positive base never gives a negative power value; base 1 is constant

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( \log_2(4 \cdot 8) \) using the product rule.
   Show the full solution

   \( \log_2 4 + \log_2 8 = 2 + 3 \). Check: \( \log_2 32 = 5 \). 5

2. Evaluate \( \log_3 27 - \log_3 3 \).
   Show the full solution

   \( 3 - 1 = 2 \), and \( \log_3 9 = 2 \). 2

3. Evaluate \( \log_5 25^3 \).
   Show the full solution

   Power rule: \( 3\log_5 25 = 3 \times 2 \). 6

4. Expand \( \log(xy) \).
   Show the full solution

   \( \log x + \log y \)

5. Condense \( \log 2 + \log 5 \).
   Show the full solution

   \( \log(2 \cdot 5) = \log 10 \). 1

6. Expand \( \log_3\dfrac{x^2\sqrt{y}}{z^4} \).
   Show the full solution

   Write \( \sqrt{y} = y^{1/2} \): \( \log_3 x^2 + \log_3 y^{1/2} - \log_3 z^4 \). Check with \( x = 3, y = 9, z = 1 \): left \( \log_3\frac{9 \cdot 3}{1} = \log_3 27 = 3 \); right \( 2 + \frac12(2) - 0 = 3 \). \( 2\log_3 x + \frac12\log_3 y - 4\log_3 z \)

7. Condense \( 3\ln x - \ln y + 2\ln 2 \) into a single logarithm.
   Show the full solution

   Powers: \( \ln x^3 - \ln y + \ln 4 \). Combine: \( \ln\dfrac{4x^3}{y} \). Check with \( x = 1, y = 2 \): left \( 0 - 0.6931 + 1.3863 = 0.6931 \); right \( \ln 2 = 0.6931 \). \( \ln\dfrac{4x^3}{y} \)

8. Evaluate \( \log_7 50 \) using change of base.
   Show the full solution

   \( \dfrac{\ln 50}{\ln 7} = \dfrac{3.91202}{1.94591} = 2.0103 \). Check: between \( \log_7 49 = 2 \) and \( \log_7 343 = 3 \), just above 2. 2.0103

9. Given \( \log_b 2 = 0.3 \) and \( \log_b 3 = 0.48 \), find \( \log_b 12 \) and \( \log_b 1.5 \).
   Show the full solution

   \( 12 = 2^2 \cdot 3 \), so \( 2(0.3) + 0.48 = 1.08 \). \( 1.5 = \dfrac32 \), so \( 0.48 - 0.3 = 0.18 \). 1.08 and 0.18

10. Show with numbers that \( \log(a + b) \ne \log a + \log b \), and state the correct statement.
    Show the full solution

    Take \( a = b = 10 \): \( \log 20 = 1.301 \) but \( \log 10 + \log 10 = 2 \). The correct statement is for a product: \( \log(ab) = \log a + \log b \), and for \( a = b = 10 \), \( \log 100 = 2 \). Adding logarithms corresponds to multiplying the arguments, never to adding them. \( \log 20 \ne 2 \); the rule is \( \log(ab) = \log a + \log b \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( 2^x = 16 \).
   Show the full solution

   \( 16 = 2^4 \). \( x = 4 \)

2. Solve \( 3^{x + 1} = 81 \).
   Show the full solution

   \( 81 = 3^4 \), so \( x + 1 = 4 \). Check: \( 3^4 = 81 \). \( x = 3 \)

3. Solve \( \log_3 x = 4 \).
   Show the full solution

   \( x = 3^4 \). \( x = 81 \)

4. Solve \( \ln x = 0 \).
   Show the full solution

   \( x = e^0 \). \( x = 1 \)

5. Solve \( 10^x = 500 \).
   Show the full solution

   \( x = \log 500 = 2.6990 \). Check: \( 10^{2.699} \approx 500 \). 2.699

6. Solve \( 4^x = 7 \).
   Show the full solution

   \( x = \dfrac{\ln 7}{\ln 4} = \dfrac{1.94591}{1.38629} = 1.4037 \). Between \( 4^1 = 4 \) and \( 4^2 = 16 \), as it must be. 1.4037

7. Solve \( 2e^{0.5x} = 10 \).
   Show the full solution

   \( e^{0.5x} = 5 \), so \( 0.5x = \ln 5 \) and \( x = 2\ln 5 = 3.2189 \). Check: \( 2e^{1.6094} = 2(5) = 10 \). 3.2189

8. Solve \( 5^{2x} = 3^{x + 1} \).
   Show the full solution

   Take logarithms: \( 2x\ln 5 = (x + 1)\ln 3 \). Collect: \( x(2\ln 5 - \ln 3) = \ln 3 \). \( x = \dfrac{1.09861}{3.21888 - 1.09861} = \dfrac{1.09861}{2.12027} = 0.5181 \). Check: \( 5^{1.0363} = 5.30 \) and \( 3^{1.5181} = 5.30 \). 0.5181

9. Solve \( \log_2(x - 1) + \log_2(x - 3) = 3 \), checking for extraneous roots.
   Show the full solution

   \( (x - 1)(x - 3) = 8 \), so \( x^2 - 4x - 5 = 0 \) and \( (x - 5)(x + 1) = 0 \). Check \( x = 5 \): \( \log_2 4 + \log_2 2 = 2 + 1 = 3 \). Correct. Check \( x = -1 \): \( x - 1 = -2 \), so the logarithm does not exist. Reject. \( x = 5 \) only

10. Solve \( \ln(x - 2) = \ln(3x - 8) \), and explain why equating the arguments is allowed.
    Show the full solution

    The logarithm is one-to-one, so equal logarithms have equal arguments: \( x - 2 = 3x - 8 \), giving \( x = 3 \). Check: \( \ln 1 = 0 \) on both sides, and both arguments equal 1, positive. Equating arguments is valid only because \( \ln \) never gives the same output for two different inputs. Without the check, a root making both arguments negative could slip in. \( x = 3 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A population of 1000 grows 5 percent a year. Find it after 6 years.
   Show the full solution

   \( 1000(1.05)^6 = 1000(1.34010) \). About 1340

2. An 80 g sample has a half-life of 10 days. How much remains after 30 days?
   Show the full solution

   30 days is 3 half-lives: \( 80 \times \frac18 \). 10 g

3. Find the doubling time at 3.5 percent compounded annually.
   Show the full solution

   \( \dfrac{\ln 2}{\ln 1.035} = \dfrac{0.69315}{0.03440} \). About 20.15 years

4. Find the value of \$1,000 after 15 years at 4 percent compounded continuously.
   Show the full solution

   \( 1000e^{0.04 \times 15} = 1000e^{0.6} \). \$1,822.12

5. Find \( k \) so that \( e^{kt} \) doubles every 5 years.
   Show the full solution

   \( e^{5k} = 2 \), so \( k = \dfrac{\ln 2}{5} \). \( k \approx 0.1386 \)

6. A drug leaves the body at 12 percent per hour. After how long is half left?
   Show the full solution

   The factor is \( 0.88 \). \( 0.88^t = 0.5 \), so \( t = \dfrac{\ln 0.5}{\ln 0.88} = \dfrac{-0.69315}{-0.12783} = 5.42 \). About 5.4 hours

7. A \$3,000 investment grows to \$4,500 in 6 years under continuous compounding. Find the rate.
   Show the full solution

   \( 3000e^{6r} = 4500 \), so \( e^{6r} = 1.5 \) and \( r = \dfrac{\ln 1.5}{6} = \dfrac{0.40547}{6} = 0.06758 \). Check: \( 3000e^{0.40547} = 4500 \). About 6.76 percent

8. Coffee at \( 200^\circ\text{F} \) cools in a \( 70^\circ\text{F} \) room following \( T = 70 + 130e^{-0.05t} \), with \( t \) in minutes. Find the temperature after 20 minutes and the time to reach \( 100^\circ\text{F} \).
   Show the full solution

   \( T(20) = 70 + 130e^{-1} = 70 + 47.83 = 117.83 \). For 100: \( 130e^{-0.05t} = 30 \), so \( e^{-0.05t} = 0.23077 \) and \( t = \dfrac{-\ln 0.23077}{0.05} = \dfrac{1.46634}{0.05} = 29.3 \). About \( 118^\circ\text{F} \), and 29.3 minutes

9. Compare \$1,000 at 6 percent simple interest with 6 percent compounded annually over 30 years.
   Show the full solution

   Simple: \( 1000(1 + 0.06 \times 30) = 2800 \). Compound: \( 1000(1.06)^{30} = 1000(5.7435) = 5743.49 \). The gap of nearly \$3,000 is interest earning interest. Simple interest is linear growth, adding a constant amount; compound interest is exponential, adding a constant percentage. \$2,800 against \$5,743

10. Show that more frequent compounding has a limit, by computing \$1 at 100 percent for one year compounded annually, monthly, daily and continuously.
    Show the full solution

    Annual: \( (1 + 1)^1 = 2 \). Monthly: \( \left(1 + \tfrac1{12}\right)^{12} = 2.6130 \). Daily: \( \left(1 + \tfrac1{365}\right)^{365} = 2.7146 \). Continuous: \( e^1 = 2.7183 \). The gains shrink: annual to monthly adds 0.613, monthly to daily adds 0.102, daily to continuous adds only 0.004. Dividing the year into finer pieces cannot raise the total above \( e \). 2, 2.613, 2.715, 2.718: the limit is \( e \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the carrying capacity of \( P(t) = \dfrac{500}{1 + 4e^{-0.2t}} \).
   Show the full solution

   The numerator is the limiting value. 500

2. Find \( P(0) \) for that model.
   Show the full solution

   \( \dfrac{500}{1 + 4} \). 100

3. Which model fits \( 4, 9, 14, 19 \) at \( t = 0, 1, 2, 3 \)?
   Show the full solution

   Differences are all 5. Linear

4. Which model fits \( 2, 6, 18, 54 \)?
   Show the full solution

   Ratios are all 3. Exponential, \( y = 2 \cdot 3^t \)

5. Describe the shape of a logistic graph.
   Show the full solution

   It starts near a floor, rises slowly, then steeply, then slows again as it approaches the ceiling. S-shaped, with a horizontal asymptote at the capacity

6. Find \( P(10) \) for \( P(t) = \dfrac{500}{1 + 4e^{-0.2t}} \).
   Show the full solution

   \( e^{-2} = 0.13534 \), denominator \( 1 + 0.54134 = 1.54134 \). \( \dfrac{500}{1.54134} = 324.4 \). About 324

7. When does that population reach 250?
   Show the full solution

   Half the capacity: \( 1 + 4e^{-0.2t} = 2 \), so \( e^{-0.2t} = \tfrac14 \) and \( t = \dfrac{\ln 4}{0.2} = 6.93 \). Check with the answer to problem 6: at \( t = 10 \) it is 324, above 250, so 250 comes earlier. Correct. About 6.93

8. A table gives \( 10, 20, 40, 80, 160 \) at \( t = 0 \) to 4. Write the model and predict \( t = 8 \).
   Show the full solution

   Ratios are 2. \( y = 10 \cdot 2^t \). At \( t = 8 \): \( 10 \cdot 256 = 2560 \). \( y = 10 \cdot 2^t \); 2560

9. Data: \( 3, 7, 13, 21, 31 \). Explain why the model is quadratic and find its equation.
   Show the full solution

   First differences 4, 6, 8, 10 are not constant; second differences 2, 2, 2 are. A quadratic \( at^2 + bt + c \) has constant second difference \( 2a \), so \( a = 1 \). With \( c = 3 \) and \( a + b + 3 = 7 \), \( b = 3 \). Check \( t = 4 \): \( 16 + 12 + 3 = 31 \). \( y = t^2 + 3t + 3 \)

10. An exponential model fits a town's population of 2000 (year 0) and 2100 (year 1) well. Predict year 100 and say whether to trust it.
    Show the full solution

    Growth factor \( 1.05 \): \( P = 2000(1.05)^t \). At \( t = 100 \): \( 2000(131.5) = 263{,}000 \), over a hundred times larger. Do not trust it. The model has no ceiling, but a town is limited by land, water and jobs. A logistic model with a realistic capacity would follow the same early data and level off. Two models can agree on the data and disagree wildly outside it, so extrapolating needs a reason beyond the fit. 263,000, which should not be trusted; the edge behavior is unrealistic

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Convert \( 90^\circ \) to radians.
   Show the full solution

   \( 90 \cdot \dfrac{\pi}{180} = \dfrac{\pi}{2} \). \( \dfrac{\pi}{2} \)

2. Convert \( \dfrac{2\pi}{3} \) to degrees.
   Show the full solution

   \( \dfrac{2 \cdot 180}{3} = 120 \). \( 120^\circ \)

3. Convert \( 225^\circ \) to radians.
   Show the full solution

   \( \dfrac{225\pi}{180} = \dfrac{5\pi}{4} \) after dividing by 45. \( \dfrac{5\pi}{4} \)

4. Give an angle coterminal with \( 50^\circ \).
   Show the full solution

   Add a turn: \( 50 + 360 = 410 \). (Subtracting gives \( -310^\circ \).) \( 410^\circ \) (or \( -310^\circ \))

5. In which quadrant does \( 160^\circ \) lie?
   Show the full solution

   Between \( 90^\circ \) and \( 180^\circ \). Quadrant II

6. Convert \( 300^\circ \) to radians and \( -\dfrac{\pi}{4} \) to a positive coterminal radian measure.
   Show the full solution

   \( \dfrac{300\pi}{180} = \dfrac{5\pi}{3} \). \( -\dfrac{\pi}{4} + 2\pi = \dfrac{7\pi}{4} \); check: \( -45^\circ + 360^\circ = 315^\circ \) and \( \dfrac{7 \cdot 180}{4} = 315 \). \( \dfrac{5\pi}{3} \) and \( \dfrac{7\pi}{4} \)

7. Find the angle in \( [0, 2\pi) \) coterminal with \( \dfrac{19\pi}{4} \).
   Show the full solution

   Subtract \( 4\pi = \dfrac{16\pi}{4} \). \( \dfrac{3\pi}{4} \)

8. Convert 2 radians to degrees and \( 72^\circ \) to radians.
   Show the full solution

   \( 2 \cdot \dfrac{180}{\pi} = 114.59^\circ \). \( 72 \cdot \dfrac{\pi}{180} = \dfrac{2\pi}{5} = 1.2566 \). Check: 2 radians is a bit more than a right angle (1.571 is \( 90^\circ \)), and \( 114.6^\circ \) is. Correct. \( 114.59^\circ \) and \( \dfrac{2\pi}{5} \approx 1.2566 \)

9. Name the quadrant of \( -\dfrac{5\pi}{6} \).
   Show the full solution

   Add \( 2\pi \): \( -\dfrac{5\pi}{6} + \dfrac{12\pi}{6} = \dfrac{7\pi}{6} \), which is \( 210^\circ \). Alternatively, going clockwise \( 150^\circ \) from the positive axis passes through IV and ends in III. Quadrant III

10. A student finds \( \sin(30) = -0.988 \) and expected 0.5. What went wrong, and how do you check?
    Show the full solution

    The calculator was in radian mode, so it computed the sine of 30 radians, not 30 degrees. 30 radians is about \( 1718.9^\circ \). Subtracting four full turns (\( 1440^\circ \)) leaves \( 278.9^\circ \), which is in quadrant IV, where sine is negative and close to \( -1 \). That is the \( -0.988 \). Always test a mode with a value you know: \( \sin(90^\circ) \) must be 1 in degree mode. Wrong mode; 30 was read as 30 radians

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the arc length for \( r = 5 \) and \( \theta = 2 \) radians.
   Show the full solution

   \( s = r\theta = 5 \cdot 2 \). 10

2. Find the sector area for \( r = 6 \) and \( \theta = \dfrac{\pi}{3} \).
   Show the full solution

   \( \tfrac12(36)\cdot\dfrac{\pi}{3} = 6\pi \). \( 6\pi \approx 18.85 \)

3. An arc of length 12 lies on a circle of radius 4. Find the angle.
   Show the full solution

   \( \theta = \dfrac{s}{r} = \dfrac{12}{4} \). 3 radians

4. Find the angular speed of a wheel turning 3 revolutions per minute.
   Show the full solution

   \( 3 \times 2\pi \). \( 6\pi \) rad/min

5. Find the arc length of a \( 30^\circ \) sector of radius 10.
   Show the full solution

   Convert first: \( 30^\circ = \dfrac{\pi}{6} \). \( s = 10 \cdot \dfrac{\pi}{6} = 5.236 \). \( \dfrac{5\pi}{3} \approx 5.24 \)

6. A sector of radius 8 cm has area 40 cm\( ^2 \). Find its angle.
   Show the full solution

   \( A = \tfrac12 r^2\theta \) gives \( \theta = \dfrac{2A}{r^2} = \dfrac{80}{64} = 1.25 \). Check: \( \tfrac12(64)(1.25) = 40 \). 1.25 radians

7. A clock's minute hand is 15 cm long. How fast does its tip move, and how far does it travel in 25 minutes?
   Show the full solution

   One turn in 60 minutes: \( \omega = \dfrac{2\pi}{60} \) rad/min. \( v = 15 \cdot \dfrac{2\pi}{60} = \dfrac{\pi}{2} = 1.571 \) cm/min. In 25 minutes the angle is \( \dfrac{25}{60}(2\pi) = \dfrac{5\pi}{6} \), so \( s = 15 \cdot \dfrac{5\pi}{6} = 12.5\pi = 39.27 \) cm. Check: \( 1.571 \times 25 = 39.27 \). 1.571 cm/min; 39.27 cm

8. A car's wheels have radius 0.3 m and the car moves at 20 m/s. Find the wheel's angular speed in rad/s and in revolutions per minute.
   Show the full solution

   \( \omega = \dfrac{v}{r} = \dfrac{20}{0.3} = 66.67 \) rad/s. Revolutions per minute: \( 66.67 \times \dfrac{60}{2\pi} = 636.6 \). 66.67 rad/s, about 637 rpm

9. Find the speed due to Earth's rotation at latitude \( 40^\circ \) north, using radius 6371 km.
   Show the full solution

   The circle at latitude \( 40^\circ \) has radius \( 6371\cos 40^\circ = 6371(0.76604) = 4880.6 \) km. \( v = \dfrac{2\pi(4880.6)}{24} = 1277.8 \) km/h. That is 77 percent of the equatorial speed \( 1668 \), and \( \cos 40^\circ = 0.766 \). About 1278 km/h

10. A student computes the arc of a \( 60^\circ \) sector with radius 10 as \( 10 \times 60 = 600 \). Explain the error and correct it.
    Show the full solution

    The formula \( s = r\theta \) needs \( \theta \) in radians. Used with 60 (degrees) it treats the angle as 60 radians, nearly ten full turns. A quick sanity check exposes it: the whole circle has circumference \( 2\pi(10) = 62.8 \), so a sixth of it cannot be 600. Correct: \( 60^\circ = \dfrac{\pi}{3} \), \( s = \dfrac{10\pi}{3} = 10.47 \), one sixth of 62.8. \( s = \dfrac{10\pi}{3} \approx 10.47 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \sin\dfrac{\pi}{2} \).
   Show the full solution

   The point is \( (0, 1) \). 1

2. Find \( \cos\pi \).
   Show the full solution

   The point is \( (-1, 0) \). \( -1 \)

3. Find \( \sin\dfrac{\pi}{6} \).
   Show the full solution

   From the first-quadrant pattern. \( \dfrac12 \)

4. Find \( \cos\dfrac{\pi}{3} \).
   Show the full solution

   The cosine values run in the reverse order of the sine values. \( \dfrac12 \)

5. Find \( \sin\pi \).
   Show the full solution

   The point is \( (-1, 0) \), and sine is the \( y \)-coordinate. 0

6. Find \( \cos\dfrac{3\pi}{4} \).
   Show the full solution

   \( \dfrac{3\pi}{4} = 135^\circ \), quadrant II, reference angle \( 45^\circ \); cosine is negative there. Check: the point \( \left(-\dfrac{\sqrt2}{2}, \dfrac{\sqrt2}{2}\right) \) is on the circle. \( -\dfrac{\sqrt2}{2} \)

7. Find \( \sin\dfrac{7\pi}{6} \).
   Show the full solution

   \( 210^\circ \), quadrant III, reference \( 30^\circ \); sine is negative there. \( -\dfrac12 \)

8. Find \( \cos\left(-\dfrac{\pi}{6}\right) \).
   Show the full solution

   Cosine is even: \( \cos\left(-\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{6} \). \( \dfrac{\sqrt3}{2} \)

9. A point \( \left(\dfrac35, y\right) \) on the unit circle lies in quadrant IV. Find \( y \), \( \sin\theta \) and \( \cos\theta \).
   Show the full solution

   \( \dfrac{9}{25} + y^2 = 1 \), so \( y^2 = \dfrac{16}{25} \) and \( y = \pm\dfrac45 \). Quadrant IV has a negative \( y \), so \( y = -\dfrac45 \). \( y = -\dfrac45 \), \( \sin\theta = -\dfrac45 \), \( \cos\theta = \dfrac35 \)

10. Explain why \( \sin^2\theta + \cos^2\theta = 1 \) for every angle, why no sine can equal 2, and what \( \sin(\theta + 2\pi) \) equals.
    Show the full solution

    The point \( (\cos\theta, \sin\theta) \) lies on the circle \( x^2 + y^2 = 1 \) by definition, so its coordinates satisfy the equation: that is the identity. Since \( y^2 \le 1 \), we have \( |y| \le 1 \), so no sine is ever 2. Adding \( 2\pi \) is a full turn and lands on the same point, so \( \sin(\theta + 2\pi) = \sin\theta \). The identity is the circle's equation; sine stays in \( [-1, 1] \); it repeats every \( 2\pi \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \sin 45^\circ \).
   Show the full solution

   Legs 1, hypotenuse \( \sqrt2 \): \( \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2} \). \( \dfrac{\sqrt2}{2} \)

2. Find \( \cos 60^\circ \).
   Show the full solution

   Adjacent to \( 60^\circ \) is the short leg 1, over hypotenuse 2. \( \dfrac12 \)

3. Find \( \tan 45^\circ \).
   Show the full solution

   The legs are equal. 1

4. In a \( 30^\circ \)-\( 60^\circ \)-\( 90^\circ \) triangle the short leg is 4. Find the other two sides.
   Show the full solution

   Hypotenuse \( 2 \times 4 = 8 \); longer leg \( 4\sqrt3 \). Check: \( 16 + 48 = 64 \). 8 and \( 4\sqrt3 \approx 6.93 \)

5. Find \( \tan 60^\circ \).
   Show the full solution

   Opposite \( \sqrt3 \), adjacent 1. \( \sqrt3 \)

6. A \( 45^\circ \)-\( 45^\circ \)-\( 90^\circ \) triangle has hypotenuse 10. Find the legs.
   Show the full solution

   Each leg is \( \dfrac{10}{\sqrt2} = 5\sqrt2 = 7.071 \). Check: \( 2(50) = 100 \). \( 5\sqrt2 \approx 7.07 \) each

7. A 6 m ladder leans at \( 65^\circ \) to the ground. How high does it reach and how far is its foot from the wall?
   Show the full solution

   Height: \( 6\sin 65^\circ = 6(0.90631) = 5.44 \) m. Foot: \( 6\cos 65^\circ = 6(0.42262) = 2.54 \) m. Check: \( 5.438^2 + 2.536^2 = 29.57 + 6.43 = 36.0 \). 5.44 m high; 2.54 m out

8. Find the acute angle whose cosine is 0.4.
   Show the full solution

   \( \cos^{-1}(0.4) = 66.42^\circ \). It lies between \( 60^\circ \) (cos 0.5) and \( 90^\circ \) (cos 0). \( 66.42^\circ \)

9. A 12 m pole casts a shadow when the Sun is \( 40^\circ \) above the horizon. Find the shadow length.
   Show the full solution

   The pole is opposite and the shadow adjacent: \( \tan 40^\circ = \dfrac{12}{s} \). \( s = \dfrac{12}{0.83910} = 14.30 \) m. Check: a lower Sun makes a longer shadow; at \( 45^\circ \) it would be 12. 14.3 m

10. Derive \( \sin 60^\circ \) from an equilateral triangle of side 2, and explain why the shortest side is opposite the smallest angle.
    Show the full solution

    The altitude is \( \sqrt3 \), opposite \( 60^\circ \), over hypotenuse 2: \( \sin 60^\circ = \dfrac{\sqrt3}{2} \). In any triangle the larger angle is opposite the longer side (the law of sines in lesson 7.2 makes this precise): \( 30^\circ \) is the smallest angle and the side opposite, 1, is the shortest. Also \( \sin 30^\circ = \cos 60^\circ \) because the two angles are complementary and each is the ratio of the same pair of sides. \( \dfrac{\sqrt3}{2} \); the smallest angle faces the shortest side

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. In which quadrant is \( 200^\circ \)?
   Show the full solution

   Between \( 180^\circ \) and \( 270^\circ \). Quadrant III

2. Find the reference angle of \( 150^\circ \).
   Show the full solution

   \( 180^\circ - 150^\circ \). \( 30^\circ \)

3. Find the reference angle of \( \dfrac{7\pi}{4} \).
   Show the full solution

   Quadrant IV: \( 2\pi - \dfrac{7\pi}{4} \). \( \dfrac{\pi}{4} \)

4. Is \( \cos 120^\circ \) positive or negative?
   Show the full solution

   Quadrant II, where only sine is positive. Negative

5. Find \( \sin 135^\circ \).
   Show the full solution

   Reference \( 45^\circ \), quadrant II, sine positive. \( \dfrac{\sqrt2}{2} \)

6. Find \( \cos 240^\circ \).
   Show the full solution

   Quadrant III, reference \( 60^\circ \), cosine negative. \( -\dfrac12 \)

7. Find \( \tan\dfrac{5\pi}{6} \).
   Show the full solution

   \( \sin = \dfrac12 \), \( \cos = -\dfrac{\sqrt3}{2} \), so \( \tan = \dfrac{1/2}{-\sqrt3/2} = -\dfrac{1}{\sqrt3} \). \( -\dfrac{\sqrt3}{3} \)

8. Find \( \sin\left(-\dfrac{2\pi}{3}\right) \).
   Show the full solution

   Add \( 2\pi \): \( \dfrac{4\pi}{3} = 240^\circ \), quadrant III, reference \( 60^\circ \), sine negative. Alternatively, sine is odd: \( -\sin\dfrac{2\pi}{3} = -\dfrac{\sqrt3}{2} \). \( -\dfrac{\sqrt3}{2} \)

9. Find all angles in \( [0, 2\pi) \) with \( \sin\theta = -\dfrac12 \) and \( \cos\theta \gt 0 \).
   Show the full solution

   Sine is negative in III and IV; cosine is positive only in IV. The reference angle is \( \dfrac{\pi}{6} \), so \( \theta = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \). Check: \( \cos\dfrac{11\pi}{6} = \dfrac{\sqrt3}{2} \gt 0 \). \( \dfrac{11\pi}{6} \)

10. If \( \theta \) is in quadrant II and \( \sin\theta = \dfrac35 \), find \( \cos\theta \) and \( \tan\theta \), and explain the sign.
    Show the full solution

    \( \cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25} \), so \( \cos\theta = \pm\dfrac45 \). In quadrant II cosine is negative: \( \cos\theta = -\dfrac45 \). \( \tan\theta = \dfrac{3/5}{-4/5} = -\dfrac34 \). The square root gives both signs; only the quadrant chooses between them. \( \cos\theta = -\dfrac45 \), \( \tan\theta = -\dfrac34 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For the point \( (3, 4) \), find \( \sin\theta \).
   Show the full solution

   \( r = 5 \), so \( \dfrac{y}{r} \). \( \dfrac45 \)

2. For \( (-5, 12) \), find \( \cos\theta \).
   Show the full solution

   \( r = 13 \), so \( \dfrac{x}{r} \). \( -\dfrac{5}{13} \)

3. For \( (-2, -2) \), find \( \tan\theta \).
   Show the full solution

   \( \dfrac{y}{x} = \dfrac{-2}{-2} \). 1

4. For \( (3, -4) \), find \( \csc\theta \).
   Show the full solution

   \( r = 5 \), and \( \csc\theta = \dfrac{r}{y} = \dfrac{5}{-4} \). \( -\dfrac54 \)

5. For \( (-1, \sqrt3) \), find \( \sec\theta \).
   Show the full solution

   \( r = \sqrt{1 + 3} = 2 \), so \( \dfrac{r}{x} = \dfrac{2}{-1} \). \( -2 \)

6. Find the angle in \( [0^\circ, 360^\circ) \) for the point \( (-3, 4) \).
   Show the full solution

   \( \tan^{-1}(-\tfrac43) = -53.13^\circ \), which is in the wrong quadrant. The point is in quadrant II, so add \( 180^\circ \). \( 126.87^\circ \)

7. For \( (2, -2\sqrt3) \), find sine, cosine and the angle in \( [0^\circ, 360^\circ) \).
   Show the full solution

   \( r = \sqrt{4 + 12} = 4 \). \( \sin\theta = -\dfrac{\sqrt3}{2} \), \( \cos\theta = \dfrac12 \). Quadrant IV, reference angle \( 60^\circ \), so \( \theta = 360^\circ - 60^\circ \). \( -\dfrac{\sqrt3}{2} \), \( \dfrac12 \), \( 300^\circ \)

8. If \( \tan\theta = \dfrac34 \) and \( \sin\theta \lt 0 \), find \( \sin\theta \) and \( \cos\theta \).
   Show the full solution

   Tangent is positive and sine negative, so quadrant III: both \( x \) and \( y \) are negative. Take \( (-4, -3) \), \( r = 5 \). \( \sin\theta = -\dfrac35 \), \( \cos\theta = -\dfrac45 \)

9. Which of the six functions are undefined at \( 0^\circ \)?
   Show the full solution

   At \( 0^\circ \) the point is \( (1, 0) \), so \( y = 0 \). The functions with \( y \) in the denominator are cosecant \( \dfrac{r}{y} \) and cotangent \( \dfrac{x}{y} \). Cosecant and cotangent

10. Explain why \( \tan^{-1}\dfrac{y}{x} \) gives the wrong angle for both \( (-3, 4) \) and \( (-3, -4) \), and give the correct angles.
    Show the full solution

    For \( (-3, 4) \): \( \dfrac{y}{x} = -\dfrac43 \), and \( \tan^{-1} \) returns \( -53.13^\circ \), a quadrant IV angle. For \( (-3, -4) \): \( \dfrac{y}{x} = \dfrac43 \), and \( \tan^{-1} \) returns \( 53.13^\circ \), a quadrant I angle. The ratio loses the signs of \( x \) and \( y \) separately, and the inverse tangent only outputs angles in \( \left(-90^\circ, 90^\circ\right) \), so it cannot reach quadrants II and III. Correct angles: \( 126.87^\circ \) and \( 180^\circ + 53.13^\circ = 233.13^\circ \). \( 126.87^\circ \) and \( 233.13^\circ \); add \( 180^\circ \) for quadrants II and III

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is \( \sin^2\theta + \cos^2\theta \)?
   Show the full solution

   It is the equation of the unit circle. 1

2. If \( \cos\theta = \dfrac35 \) and \( \theta \) is in quadrant I, find \( \sin\theta \).
   Show the full solution

   \( \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25} \); positive in quadrant I. \( \dfrac45 \)

3. If \( \sin\theta = \dfrac{5}{13} \) and \( \theta \) is in quadrant II, find \( \cos\theta \).
   Show the full solution

   \( \cos^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169} \); negative in quadrant II. \( -\dfrac{12}{13} \)

4. If \( \sin\theta = -\dfrac35 \) and \( \cos\theta = \dfrac45 \), find \( \tan\theta \).
   Show the full solution

   \( \dfrac{-3/5}{4/5} \). \( -\dfrac34 \)

5. Simplify \( 1 - \sin^2\theta \).
   Show the full solution

   Rearrange the Pythagorean identity. \( \cos^2\theta \)

6. If \( \tan\theta = -2 \) and \( \theta \) is in quadrant IV, find \( \sec\theta \), \( \cos\theta \) and \( \sin\theta \).
   Show the full solution

   \( \sec^2\theta = 1 + 4 = 5 \), so \( \sec\theta = \pm\sqrt5 \); positive in quadrant IV. \( \cos\theta = \dfrac{1}{\sqrt5} \), and \( \sin\theta = \tan\theta\cos\theta = -\dfrac{2}{\sqrt5} \). Check: \( \dfrac15 + \dfrac45 = 1 \). \( \sqrt5 \), \( \dfrac{\sqrt5}{5} \), \( -\dfrac{2\sqrt5}{5} \)

7. Simplify \( \sin\theta\sec\theta \).
   Show the full solution

   \( \sin\theta \cdot \dfrac{1}{\cos\theta} = \tan\theta \). Check at \( 30^\circ \): \( 0.5 \times 1.1547 = 0.5774 = \tan 30^\circ \). \( \tan\theta \)

8. If \( \csc\theta = -\dfrac53 \) and \( \theta \) is in quadrant IV, find \( \cos\theta \) and \( \tan\theta \).
   Show the full solution

   \( \sin\theta = -\dfrac35 \). \( \cos^2\theta = \dfrac{16}{25} \), positive in quadrant IV: \( \cos\theta = \dfrac45 \). \( \tan\theta = -\dfrac34 \). \( \dfrac45 \) and \( -\dfrac34 \)

9. Simplify \( \dfrac{\sec^2 x - 1}{\sec^2 x} \).
   Show the full solution

   \( \sec^2 x - 1 = \tan^2 x \), so the expression is \( \dfrac{\tan^2 x}{\sec^2 x} = \dfrac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x = \sin^2 x \). Check at \( x = 60^\circ \): \( \dfrac{4 - 1}{4} = \dfrac34 \) and \( \sin^2 60^\circ = \dfrac34 \). \( \sin^2 x \)

10. Derive \( 1 + \cot^2\theta = \csc^2\theta \), and say where it fails.
    Show the full solution

    Divide \( \sin^2\theta + \cos^2\theta = 1 \) by \( \sin^2\theta \): \( 1 + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta} \), that is \( 1 + \cot^2\theta = \csc^2\theta \). The division requires \( \sin\theta \ne 0 \), so the identity holds except at multiples of \( \pi \), where cotangent and cosecant are undefined, consistent with lesson 4.6. Check at \( 45^\circ \): \( 1 + 1 = 2 = (\sqrt2)^2 \). Valid wherever \( \sin\theta \ne 0 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the maximum value of \( y = \sin x \)?
   Show the full solution

   The \( y \)-coordinate on the unit circle never exceeds 1. 1

2. State the range of \( y = \cos x \).
   Show the full solution

   \( [-1, 1] \)

3. State the period of \( y = \sin x \).
   Show the full solution

   One full turn. \( 2\pi \)

4. Find all \( x \) in \( [0, 2\pi] \) with \( \sin x = 0 \).
   Show the full solution

   The points on the \( x \)-axis. \( 0, \pi, 2\pi \)

5. Evaluate \( \cos\pi \).
   Show the full solution

   The point is \( (-1, 0) \). \( -1 \)

6. Find all \( x \) in \( [0, 2\pi] \) with \( \cos x = 0 \).
   Show the full solution

   The points \( (0, 1) \) and \( (0, -1) \). \( \dfrac{\pi}{2} \) and \( \dfrac{3\pi}{2} \)

7. Find all \( x \) in \( [-2\pi, 2\pi] \) with \( \sin x = 1 \).
   Show the full solution

   The maximum occurs at \( \dfrac{\pi}{2} \) and every \( 2\pi \) from it: \( \dfrac{\pi}{2} \) and \( \dfrac{\pi}{2} - 2\pi = -\dfrac{3\pi}{2} \). Check: \( \sin\left(-\dfrac{3\pi}{2}\right) = 1 \). \( x = \dfrac{\pi}{2} \) and \( x = -\dfrac{3\pi}{2} \)

8. Solve \( \sin x = -\dfrac12 \) on \( [0, 2\pi] \).
   Show the full solution

   Negative sine: quadrants III and IV, reference angle \( \dfrac{\pi}{6} \). \( x = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6} \) and \( x = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \). \( \dfrac{7\pi}{6} \) and \( \dfrac{11\pi}{6} \)

9. Is \( \sin x \) even or odd? Explain using the graph and test with \( x = \dfrac{\pi}{6} \).
   Show the full solution

   Odd: the graph is unchanged by a half-turn about the origin. \( \sin\left(-\dfrac{\pi}{6}\right) = -\dfrac12 = -\sin\dfrac{\pi}{6} \). Odd, so \( \sin(-x) = -\sin x \)

10. Show that \( \cos x = \sin\left(x + \dfrac{\pi}{2}\right) \) at \( x = \dfrac{\pi}{6} \), and say what this means for the graphs.
    Show the full solution

    \( \cos\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} \). \( \sin\left(\dfrac{\pi}{6} + \dfrac{\pi}{2}\right) = \sin\dfrac{2\pi}{3} = \dfrac{\sqrt3}{2} \). Equal. The cosine curve is the sine curve moved left by \( \dfrac{\pi}{2} \). Cosine is sine shifted left by \( \dfrac{\pi}{2} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the amplitude of \( y = -4\sin x \).
   Show the full solution

   Amplitude is a distance, so take the absolute value. 4

2. Find the period of \( y = \sin 3x \).
   Show the full solution

   \( \dfrac{2\pi}{3} \). \( \dfrac{2\pi}{3} \)

3. Find the midline of \( y = \cos x + 5 \).
   Show the full solution

   \( y = 5 \)

4. Find the maximum and minimum of \( y = 2\sin x - 1 \).
   Show the full solution

   \( -1 + 2 = 1 \) and \( -1 - 2 = -3 \). 1 and \( -3 \)

5. Find the period of \( y = \cos\dfrac{x}{2} \).
   Show the full solution

   \( B = \dfrac12 \), so \( 2\pi \div \dfrac12 = 4\pi \). \( 4\pi \)

6. State the phase shift of \( y = \sin\left(x - \dfrac{\pi}{3}\right) \).
   Show the full solution

   The form \( x - C \) with \( C = \dfrac{\pi}{3} \). \( \dfrac{\pi}{3} \) to the right

7. Find the period and phase shift of \( y = 3\cos(2x - \pi) \).
   Show the full solution

   \( 2x - \pi = 2\left(x - \dfrac{\pi}{2}\right) \). Period \( \dfrac{2\pi}{2} = \pi \); shift \( \dfrac{\pi}{2} \) to the right. Check: at \( x = \dfrac{\pi}{2} \) the argument is 0 and the value is the maximum 3. Period \( \pi \); shift \( \dfrac{\pi}{2} \) right

8. List the five key points of \( y = -2\cos x + 1 \) over one period from 0 to \( 2\pi \).
   Show the full solution

   Cosine's values \( 1, 0, -1, 0, 1 \) become \( -2(1) + 1 = -1 \), \( 1 \), \( 3 \), \( 1 \), \( -1 \). \( (0, -1), \left(\tfrac{\pi}{2}, 1\right), (\pi, 3), \left(\tfrac{3\pi}{2}, 1\right), (2\pi, -1) \)

9. Find the period of \( y = \sin(\pi x) \) and the maximum value, and where the first maximum occurs.
   Show the full solution

   \( B = \pi \), period \( \dfrac{2\pi}{\pi} = 2 \). The maximum is 1, at a quarter period: \( x = 0.5 \). Check: \( \sin\dfrac{\pi}{2} = 1 \). Period 2; maximum 1 at \( x = 0.5 \)

10. A student says \( y = \sin(2x + \pi) \) has a phase shift of \( \pi \) to the left. Correct them.
    Show the full solution

    Factor the coefficient of \( x \): \( 2x + \pi = 2\left(x + \dfrac{\pi}{2}\right) \). The shift is \( \dfrac{\pi}{2} \) to the left. Check: the graph starts a cycle where the argument is 0, at \( x = -\dfrac{\pi}{2} \), and \( \sin(2 \cdot -\dfrac{\pi}{2} + \pi) = \sin 0 = 0 \). At \( x = -\pi \) the argument is \( -\pi \), not a cycle start. \( \dfrac{\pi}{2} \) left

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the amplitude of a wave with maximum 9 and minimum \( -3 \).
   Show the full solution

   \( \dfrac{9 - (-3)}{2} \). 6

2. Find the midline of that wave.
   Show the full solution

   \( \dfrac{9 + (-3)}{2} \). 3

3. Consecutive maxima occur at \( x = 2 \) and \( x = 10 \). Find the period.
   Show the full solution

   \( 10 - 2 \). 8

4. Find \( B \) for a period of 8.
   Show the full solution

   \( B = \dfrac{2\pi}{8} \). \( \dfrac{\pi}{4} \)

5. Write a cosine with amplitude 3, period \( 2\pi \), midline 0, maximum at \( x = 0 \).
   Show the full solution

   \( B = 1 \). \( y = 3\cos x \)

6. Write the equation: amplitude 5, midline 1, period \( \pi \), maximum at \( x = 0 \).
   Show the full solution

   \( B = \dfrac{2\pi}{\pi} = 2 \). Check: \( y(0) = 5 + 1 = 6 \), \( y\left(\dfrac{\pi}{2}\right) = 5\cos\pi + 1 = -4 \). \( y = 5\cos 2x + 1 \)

7. A wave has its maximum 10 at \( x = 0 \) and its next minimum 2 at \( x = 6 \). Write its equation as a cosine.
   Show the full solution

   Amplitude 4, midline 6, half period 6 so period 12, \( B = \dfrac{\pi}{6} \). Check at \( x = 6 \): \( 4\cos\pi + 6 = 2 \). \( y = 4\cos\left(\dfrac{\pi x}{6}\right) + 6 \)

8. Show that \( 3\sin\left(x - \dfrac{\pi}{2}\right) = -3\cos x \).
   Show the full solution

   Shifting sine right by a quarter period gives negative cosine. Check at \( x = 0 \): \( 3\sin\left(-\dfrac{\pi}{2}\right) = -3 \) and \( -3\cos 0 = -3 \). At \( x = \pi \): \( 3\sin\dfrac{\pi}{2} = 3 \) and \( -3\cos\pi = 3 \). They describe the same curve

9. A sinusoid rises through its midline \( -1 \) at \( x = 1 \) and first reaches its maximum 1 at \( x = 2.5 \). Write its equation.
   Show the full solution

   Amplitude 2, midline \( -1 \). The quarter period is 1.5, so the period is 6 and \( B = \dfrac{\pi}{3} \). It starts at the midline going up at \( x = 1 \): \( y = 2\sin\left(\dfrac{\pi}{3}(x - 1)\right) - 1 \). Check at \( x = 2.5 \): \( 2\sin\dfrac{\pi}{2} - 1 = 1 \) ✓. \( y = 2\sin\left(\dfrac{\pi}{3}(x - 1)\right) - 1 \)

10. Give a cosine and a sine equation for the same graph (max 7 at \( x = 1 \), min \( -1 \) at \( x = 4 \)) and show they agree at \( x = 2 \).
    Show the full solution

    Cosine: \( 4\cos\left(\dfrac{\pi}{3}(x - 1)\right) + 3 \). At \( x = 2 \): \( 4\cos\dfrac{\pi}{3} + 3 = 4(0.5) + 3 = 5 \). Sine: \( 4\sin\left(\dfrac{\pi}{3}(x + 0.5)\right) + 3 \). At \( x = 2 \): \( 4\sin\dfrac{5\pi}{6} + 3 = 4(0.5) + 3 = 5 \). They agree because sine and cosine are the same wave shifted a quarter period. Both give 5

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Using the Ferris wheel of the example, find the height at \( t = 5 \).
   Show the full solution

   \( -20\cos\dfrac{\pi}{4} + 22 = 22 - 14.142 \). 7.86 m

2. What is the greatest height the rider reaches?
   Show the full solution

   Midline plus amplitude: \( 22 + 20 \). 42 m

3. What is the period of the wheel?
   Show the full solution

   \( \dfrac{2\pi}{\pi/20} \). 40 s

4. For \( d(t) = 6 + 2.5\cos\left(\dfrac{\pi}{6}(t - 3)\right) \), find \( d(3) \).
   Show the full solution

   The cosine is \( \cos 0 = 1 \). 8.5 m

5. Find \( d(9) \).
   Show the full solution

   \( \cos\pi = -1 \), so \( 6 - 2.5 \). 3.5 m

6. When is the rider first 32 m up?
   Show the full solution

   \( \cos\dfrac{\pi t}{20} = -\dfrac12 \), so \( \dfrac{\pi t}{20} = \dfrac{2\pi}{3} \) and \( t = 13.33 \) s. Check: \( 22 + 10 = 32 \). 13.33 s

7. For how many hours per cycle is the harbor deeper than 7 m?
   Show the full solution

   \( \cos\theta \gt 0.4 \) gives \( |\theta| \lt 1.1593 \), so the window is \( 2 \times 2.214 = 4.43 \) hours. About 4.43 hours

8. Monthly average temperature is \( T(m) = 14 + 10\sin\left(\dfrac{\pi}{6}(m - 4)\right) \) degrees, with \( m = 1 \) for January. Find \( T(1) \), \( T(7) \) and \( T(10) \).
   Show the full solution

   \( T(1) = 14 + 10\sin\left(-\dfrac{\pi}{2}\right) = 4 \). \( T(7) = 14 + 10\sin\dfrac{\pi}{2} = 24 \). \( T(10) = 14 + 10\sin\pi = 14 \). The coldest month is January and the warmest July, a half-period apart. 4, 24 and 14 degrees

9. A wheel of radius 15 m has its center 17 m above the ground and turns once in 30 s, starting at the bottom. Write \( h(t) \) and find the height at \( t = 10 \).
   Show the full solution

   \( B = \dfrac{2\pi}{30} = \dfrac{\pi}{15} \), so \( h(t) = -15\cos\dfrac{\pi t}{15} + 17 \). \( h(10) = -15\cos\dfrac{2\pi}{3} + 17 = 7.5 + 17 = 24.5 \). Check \( h(0) = 2 \), the bottom. \( h(t) = -15\cos\frac{\pi t}{15} + 17 \); 24.5 m

10. At what times in \( 0 \le t \le 12 \) is the harbor at its mean depth of 6 m? Explain the significance.
    Show the full solution

    \( 6 + 2.5\cos\theta = 6 \) means \( \cos\theta = 0 \), so \( \theta = \dfrac{\pi}{6}(t - 3) = \pm\dfrac{\pi}{2} \), giving \( t - 3 = \pm 3 \), so \( t = 0 \) or \( t = 6 \), and again at \( t = 12 \). These are the midline crossings, a quarter cycle from high tide, when the water is moving fastest: rising or falling at its greatest rate. At the highs and lows the depth is momentarily steady. \( t = 0, 6, 12 \); the depth changes fastest there

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the period of \( y = \tan x \).
   Show the full solution

   \( \pi \)

2. Name the vertical asymptotes of \( y = \tan x \) between \( -\dfrac{\pi}{2} \) and \( \dfrac{\pi}{2} \) inclusive.
   Show the full solution

   Where cosine is zero. \( x = -\dfrac{\pi}{2} \) and \( x = \dfrac{\pi}{2} \)

3. Evaluate \( \tan\dfrac{\pi}{4} \).
   Show the full solution

   Sine equals cosine there. 1

4. State the range of \( y = \sec x \).
   Show the full solution

   \( \sec x = \dfrac{1}{\cos x} \) and \( |\cos x| \le 1 \). \( (-\infty, -1] \cup [1, \infty) \)

5. Evaluate \( \sec 0 \).
   Show the full solution

   \( \dfrac{1}{\cos 0} \). 1

6. Find the period of \( y = \tan 3x \) and its first positive asymptote.
   Show the full solution

   Period \( \dfrac{\pi}{3} \). Asymptote: \( 3x = \dfrac{\pi}{2} \), so \( x = \dfrac{\pi}{6} \). Period \( \dfrac{\pi}{3} \); first asymptote \( x = \dfrac{\pi}{6} \)

7. Evaluate \( \csc\dfrac{\pi}{6} \).
   Show the full solution

   \( \dfrac{1}{\sin(\pi/6)} = \dfrac{1}{1/2} \). 2

8. Find the minimum positive value of \( y = 2\sec x \) and where it occurs.
   Show the full solution

   \( \sec x \ge 1 \) on the upper branches, with equality at \( x = 0 \), so \( 2\sec x \ge 2 \). The value on the lower branches is at most \( -2 \). 2, at \( x = 0 \)

9. Find the asymptotes and zeros of \( y = \cot x \) on \( (0, 2\pi) \).
   Show the full solution

   \( \cot x = \dfrac{\cos x}{\sin x} \). Asymptote where sine is 0: \( x = \pi \) (and the ends 0 and \( 2\pi \)). Zeros where cosine is 0: \( \dfrac{\pi}{2} \) and \( \dfrac{3\pi}{2} \). Check: \( \cot\dfrac{\pi}{2} = 0 \). Asymptote \( x = \pi \); zeros \( \dfrac{\pi}{2}, \dfrac{3\pi}{2} \)

10. Explain why tangent and secant share their asymptotes, and why secant never takes values between \( -1 \) and 1.
    Show the full solution

    Both have \( \cos x \) in the denominator: \( \tan x = \dfrac{\sin x}{\cos x} \) and \( \sec x = \dfrac{1}{\cos x} \). They fail wherever \( \cos x = 0 \). Since \( -1 \le \cos x \le 1 \), the reciprocal of a number of size at most 1 has size at least 1: if \( |\cos x| = 0.5 \) then \( |\sec x| = 2 \). Small cosine gives a huge secant, and a cosine of exactly \( \pm 1 \) gives \( \pm 1 \), the closest to zero it can be. Same denominator; the reciprocal of a number in \( [-1, 1] \) has size at least 1

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( \sin^{-1}0 \).
   Show the full solution

   0

2. Evaluate \( \cos^{-1}1 \).
   Show the full solution

   0

3. Evaluate \( \tan^{-1}1 \).
   Show the full solution

   \( \dfrac{\pi}{4} \)

4. Evaluate \( \sin^{-1}\dfrac{\sqrt2}{2} \).
   Show the full solution

   \( \dfrac{\pi}{4} \)

5. Evaluate \( \cos^{-1}0 \).
   Show the full solution

   The angle in \( [0, \pi] \) with cosine 0. \( \dfrac{\pi}{2} \)

6. Evaluate \( \cos^{-1}\left(-\dfrac{\sqrt3}{2}\right) \).
   Show the full solution

   Quadrant II, reference angle \( \dfrac{\pi}{6} \): \( \pi - \dfrac{\pi}{6} \). Check: \( \cos\dfrac{5\pi}{6} = -\dfrac{\sqrt3}{2} \). \( \dfrac{5\pi}{6} \)

7. Evaluate \( \sin^{-1}\left(-\dfrac12\right) \).
   Show the full solution

   Negative, so in \( \left[-\dfrac{\pi}{2}, 0\right) \); reference angle \( \dfrac{\pi}{6} \). \( -\dfrac{\pi}{6} \)

8. Evaluate \( \sin^{-1}\left(\sin\dfrac{7\pi}{6}\right) \).
   Show the full solution

   \( \sin\dfrac{7\pi}{6} = -\dfrac12 \). The angle in the range with sine \( -\dfrac12 \) is \( -\dfrac{\pi}{6} \). \( -\dfrac{\pi}{6} \), not \( \dfrac{7\pi}{6} \)

9. Find the domain of \( f(x) = \sin^{-1}(2x - 1) \).
   Show the full solution

   Need \( -1 \le 2x - 1 \le 1 \); add 1: \( 0 \le 2x \le 2 \); divide: \( 0 \le x \le 1 \). Check: \( x = 1 \) gives \( \sin^{-1}1 = \dfrac{\pi}{2} \); \( x = 2 \) gives \( \sin^{-1}3 \), undefined. \( [0, 1] \)

10. Solve \( \sin x = 0.3 \) for all \( x \) in \( [0, 2\pi] \), and explain why the calculator gives only one.
    Show the full solution

    \( \sin^{-1}(0.3) = 0.3047 \), in quadrant I. Sine is also positive in quadrant II: \( \pi - 0.3047 = 2.8369 \). Check: \( \sin 2.8369 = 0.3 \). The calculator returns the one value in the restricted range \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \); the rest of the solutions come from the symmetry of the circle. \( x = 0.3047 \) and \( x = 2.8369 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( \sin\left(\sin^{-1}\dfrac12\right) \).
   Show the full solution

   A function and its inverse cancel: the input \( \dfrac12 \) is in \( [-1, 1] \). \( \dfrac12 \)

2. Evaluate \( \cos\left(\cos^{-1}(-0.3)\right) \).
   Show the full solution

   \( -0.3 \) is in \( [-1, 1] \). \( -0.3 \)

3. Evaluate \( \cos\left(\sin^{-1}\dfrac35\right) \).
   Show the full solution

   Triangle with opposite 3, hypotenuse 5, adjacent 4. \( \dfrac45 \)

4. Evaluate \( \tan\left(\sin^{-1}\dfrac{5}{13}\right) \).
   Show the full solution

   Opposite 5, hypotenuse 13, adjacent 12. \( \dfrac{5}{12} \)

5. Evaluate \( \sin^{-1}\left(\sin\dfrac{\pi}{6}\right) \).
   Show the full solution

   \( \dfrac{\pi}{6} \) is in the range. \( \dfrac{\pi}{6} \)

6. Evaluate \( \sin^{-1}\left(\sin\dfrac{3\pi}{4}\right) \).
   Show the full solution

   \( \sin\dfrac{3\pi}{4} = \dfrac{\sqrt2}{2} \); the angle in the range with that sine is \( \dfrac{\pi}{4} \). \( \dfrac{\pi}{4} \)

7. Evaluate \( \cos^{-1}\left(\cos\left(-\dfrac{\pi}{3}\right)\right) \).
   Show the full solution

   \( \cos\left(-\dfrac{\pi}{3}\right) = \dfrac12 \); the angle in \( [0, \pi] \) is \( \dfrac{\pi}{3} \). \( \dfrac{\pi}{3} \)

8. Evaluate \( \sin\left(\cos^{-1}\left(-\dfrac{5}{13}\right)\right) \).
   Show the full solution

   Quadrant II; magnitudes 5, 12, 13; sine is positive. \( \dfrac{12}{13} \)

9. Evaluate \( \tan\left(\cos^{-1}\left(-\dfrac35\right)\right) \).
   Show the full solution

   Quadrant II: \( \sin = \dfrac45 \), \( \cos = -\dfrac35 \). \( -\dfrac43 \)

10. Write \( \cos(\tan^{-1}x) \) as an algebraic expression and verify it at \( x = -\sqrt3 \).
    Show the full solution

    Triangle with opposite \( x \), adjacent 1, hypotenuse \( \sqrt{1 + x^2} \): the cosine is \( \dfrac{1}{\sqrt{1 + x^2}} \). At \( x = -\sqrt3 \): \( \tan^{-1}(-\sqrt3) = -\dfrac{\pi}{3} \) and \( \cos\left(-\dfrac{\pi}{3}\right) = \dfrac12 \); the formula gives \( \dfrac{1}{\sqrt{1 + 3}} = \dfrac12 \) ✓. \( \dfrac{1}{\sqrt{1 + x^2}} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \sin x\csc x \).
   Show the full solution

   \( \csc x = \dfrac{1}{\sin x} \), so the product is 1 (for \( \sin x \ne 0 \)). 1

2. Simplify \( \tan x\cos x \).
   Show the full solution

   \( \sin x \)

3. Simplify \( 1 - \sin^2 x \).
   Show the full solution

   Rearranging \( \sin^2 x + \cos^2 x = 1 \). \( \cos^2 x \)

4. Simplify \( \sec^2 x - \tan^2 x \).
   Show the full solution

   From \( 1 + \tan^2 x = \sec^2 x \). 1

5. Simplify \( \cot x\sin x \).
   Show the full solution

   \( \dfrac{\cos x}{\sin x}\cdot\sin x \). \( \cos x \)

6. Simplify \( \dfrac{1 - \cos^2 x}{\sin x} \).
   Show the full solution

   The numerator is \( \sin^2 x \), so the quotient is \( \sin x \), for \( \sin x \ne 0 \). \( \sin x \)

7. Simplify \( \cos^2 x(1 + \tan^2 x) \).
   Show the full solution

   \( 1 + \tan^2 x = \sec^2 x \), so \( \cos^2 x\sec^2 x = 1 \). Check at \( x = \dfrac{\pi}{3} \): \( \dfrac14(1 + 3) = 1 \). 1

8. Simplify \( \dfrac{\sin^2 x}{1 - \cos x} \).
   Show the full solution

   \( \sin^2 x = (1 - \cos x)(1 + \cos x) \); cancel \( 1 - \cos x \). \( 1 + \cos x \)

9. Write \( \sec x - \cos x \) as a product and check at \( x = \dfrac{\pi}{3} \).
   Show the full solution

   \( \dfrac{1 - \cos^2 x}{\cos x} = \sin x\tan x \). At \( \dfrac{\pi}{3} \): \( \sec = 2 \), \( \cos = 0.5 \), difference 1.5; and \( \dfrac{\sqrt3}{2}(\sqrt3) = 1.5 \). \( \sin x\tan x \)

10. Show that \( \sin x + \cos x = 1 \) is not an identity.
    Show the full solution

    At \( x = \dfrac{\pi}{4} \) the left side is \( \dfrac{\sqrt2}{2} + \dfrac{\sqrt2}{2} = \sqrt2 \approx 1.414 \), not 1. It is true at \( x = 0 \), which shows why one matching example proves nothing. It fails at \( \dfrac{\pi}{4} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Prove \( \sin x\cot x = \cos x \).
   Show the full solution

   \( \sin x\cdot\dfrac{\cos x}{\sin x} = \cos x \). ∎ Left side equals right side

2. Prove \( \tan x + \cot x = \sec x\csc x \).
   Show the full solution

   \( \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x} = \sec x\csc x \). ∎ Combined over a common denominator

3. Prove \( (1 - \sin x)(1 + \sin x) = \cos^2 x \).
   Show the full solution

   \( 1 - \sin^2 x = \cos^2 x \). ∎ Difference of squares

4. Prove \( (\sin x + \cos x)^2 = 1 + 2\sin x\cos x \).
   Show the full solution

   Expand: \( \sin^2 x + 2\sin x\cos x + \cos^2 x = 1 + 2\sin x\cos x \). ∎ Pythagorean identity

5. Prove \( (\csc x - \cot x)(\csc x + \cot x) = 1 \).
   Show the full solution

   Difference of squares: \( \csc^2 x - \cot^2 x = 1 \), from \( 1 + \cot^2 x = \csc^2 x \). ∎ Equal to 1

6. Prove \( (1 - \cos^2 x)(1 + \cot^2 x) = 1 \).
   Show the full solution

   \( \sin^2 x\cdot\csc^2 x = \sin^2 x\cdot\dfrac{1}{\sin^2 x} = 1 \). ∎ Equal to 1

7. Prove \( \dfrac{\sin x}{1 + \cos x} + \dfrac{1 + \cos x}{\sin x} = 2\csc x \).
   Show the full solution

   Common denominator: \( \dfrac{\sin^2 x + (1 + \cos x)^2}{\sin x(1 + \cos x)} \). Numerator: \( \sin^2 x + 1 + 2\cos x + \cos^2 x = 2 + 2\cos x = 2(1 + \cos x) \). The factor \( 1 + \cos x \) cancels: \( \dfrac{2}{\sin x} = 2\csc x \). ∎ Left side equals \( 2\csc x \)

8. Prove \( \dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = 2\sec^2 x \).
   Show the full solution

   \( \dfrac{(1 + \sin x) + (1 - \sin x)}{1 - \sin^2 x} = \dfrac{2}{\cos^2 x} = 2\sec^2 x \). ∎ Left side equals \( 2\sec^2 x \)

9. Prove \( \tan^2 x - \sin^2 x = \tan^2 x\sin^2 x \).
   Show the full solution

   \( \dfrac{\sin^2 x}{\cos^2 x} - \sin^2 x = \sin^2 x\left(\dfrac{1}{\cos^2 x} - 1\right) = \sin^2 x\cdot\dfrac{1 - \cos^2 x}{\cos^2 x} = \sin^2 x\cdot\dfrac{\sin^2 x}{\cos^2 x} = \sin^2 x\tan^2 x \). ∎ Check at \( x = \dfrac{\pi}{3} \): \( 3 - 0.75 = 2.25 \) and \( 0.75 \times 3 = 2.25 \). Both sides equal \( \tan^2 x\sin^2 x \)

10. Explain why cross-multiplying is not allowed in a proof, using the claim \( \sin^2 x = 1 - \cos x \).
    Show the full solution

    The claim is false: at \( x = \dfrac{\pi}{3} \), \( \sin^2 x = 0.75 \) while \( 1 - \cos x = 0.5 \). Yet manipulating both sides can seem to "work": multiply both sides by 0 and get \( 0 = 0 \). Steps that are true for a false claim prove nothing about the claim, since they are not reversible. A proof must start from a side we know is defined and reach the other by steps that are valid on their own. Cross-multiplying assumes what is to be proved

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the exact value of \( \sin 75^\circ \).
   Show the full solution

   \( \sin(45^\circ + 30^\circ) = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt2}{2}\cdot\dfrac12 \). \( \dfrac{\sqrt6 + \sqrt2}{4} \approx 0.9659 \)

2. Find the exact value of \( \cos 15^\circ \).
   Show the full solution

   \( \cos(45^\circ - 30^\circ) = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt2}{2}\cdot\dfrac12 \). \( \dfrac{\sqrt6 + \sqrt2}{4} \approx 0.9659 \)

3. Find the exact value of \( \tan 15^\circ \).
   Show the full solution

   \( \tan(45^\circ - 30^\circ) = \dfrac{1 - \frac{\sqrt3}{3}}{1 + \frac{\sqrt3}{3}} = \dfrac{3 - \sqrt3}{3 + \sqrt3} \). Multiply by the conjugate \( 3 - \sqrt3 \): \( \dfrac{(3 - \sqrt3)^2}{9 - 3} = \dfrac{12 - 6\sqrt3}{6} = 2 - \sqrt3 \approx 0.2679 \). \( 2 - \sqrt3 \)

4. Find the exact value of \( \sin 105^\circ \).
   Show the full solution

   \( \sin(60^\circ + 45^\circ) = \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2} + \dfrac12\cdot\dfrac{\sqrt2}{2} \). \( \dfrac{\sqrt6 + \sqrt2}{4} \)

5. Simplify \( \sin 40^\circ\cos 20^\circ + \cos 40^\circ\sin 20^\circ \).
   Show the full solution

   This is \( \sin(40^\circ + 20^\circ) = \sin 60^\circ \). \( \dfrac{\sqrt3}{2} \)

6. Simplify \( \cos 70^\circ\cos 20^\circ - \sin 70^\circ\sin 20^\circ \).
   Show the full solution

   This is \( \cos(70^\circ + 20^\circ) = \cos 90^\circ \). 0

7. Find the exact value of \( \cos\dfrac{7\pi}{12} \).
   Show the full solution

   \( \dfrac{7\pi}{12} = \dfrac{\pi}{3} + \dfrac{\pi}{4} \) (that is \( 105^\circ \)). \( \cos\dfrac{\pi}{3}\cos\dfrac{\pi}{4} - \sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4} = \dfrac12\cdot\dfrac{\sqrt2}{2} - \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2} = \dfrac{\sqrt2 - \sqrt6}{4} \approx -0.2588 \). Negative, as quadrant II requires. \( \dfrac{\sqrt2 - \sqrt6}{4} \)

8. If \( \sin A = \dfrac45 \) and \( \sin B = \dfrac{5}{13} \), both in quadrant I, find \( \sin(A + B) \).
   Show the full solution

   \( \cos A = \dfrac35 \) and \( \cos B = \dfrac{12}{13} \). \( \dfrac45\cdot\dfrac{12}{13} + \dfrac35\cdot\dfrac{5}{13} = \dfrac{48 + 15}{65} \). \( \dfrac{63}{65} \)

9. Use the difference formula to prove \( \cos\left(\dfrac{\pi}{2} - x\right) = \sin x \).
   Show the full solution

   \( \cos\dfrac{\pi}{2}\cos x + \sin\dfrac{\pi}{2}\sin x = 0\cdot\cos x + 1\cdot\sin x = \sin x \). ∎ The cofunction identity follows from the formula

10. Show with \( A = B = \dfrac{\pi}{6} \) that \( \sin(A + B) \ne \sin A + \sin B \), and find the correct value.
    Show the full solution

    \( \sin A + \sin B = \dfrac12 + \dfrac12 = 1 \), but \( \sin(A + B) = \sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2} \approx 0.866 \). The correct formula gives \( \dfrac12\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt3}{2}\cdot\dfrac12 = \dfrac{\sqrt3}{2} \) ✓. \( 1 \ne 0.866 \); the sum formula gives \( \dfrac{\sqrt3}{2} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( \sin x = \dfrac35 \) and \( \cos x = \dfrac45 \), find \( \sin 2x \).
   Show the full solution

   \( 2\cdot\dfrac35\cdot\dfrac45 \). \( \dfrac{24}{25} \)

2. If \( \cos x = \dfrac35 \), find \( \cos 2x \).
   Show the full solution

   \( 2\cos^2 x - 1 = 2\cdot\dfrac{9}{25} - 1 = \dfrac{18 - 25}{25} \). \( -\dfrac{7}{25} \)

3. Use \( \sin 2x \) with \( x = 30^\circ \) to find \( \sin 60^\circ \).
   Show the full solution

   \( 2\sin 30^\circ\cos 30^\circ = 2\cdot\dfrac12\cdot\dfrac{\sqrt3}{2} \). \( \dfrac{\sqrt3}{2} \)

4. If \( \tan x = \dfrac12 \), find \( \tan 2x \).
   Show the full solution

   \( \dfrac{2(1/2)}{1 - 1/4} = \dfrac{1}{3/4} \). \( \dfrac43 \)

5. If \( \sin x = \dfrac13 \), find \( \cos 2x \).
   Show the full solution

   Use \( 1 - 2\sin^2 x = 1 - \dfrac29 \). \( \dfrac79 \)

6. Find the exact value of \( \cos 15^\circ \) using the half-angle formula.
   Show the full solution

   \( \sqrt{\dfrac{1 + \cos 30^\circ}{2}} = \sqrt{\dfrac{2 + \sqrt3}{4}} = \dfrac{\sqrt{2 + \sqrt3}}{2} \approx 0.9659 \). This matches \( \dfrac{\sqrt6 + \sqrt2}{4} \) from the last lesson. \( \dfrac{\sqrt{2 + \sqrt3}}{2} \)

7. Find the exact value of \( \sin\dfrac{\pi}{8} \).
   Show the full solution

   \( \dfrac{\pi}{8} = \dfrac12\cdot\dfrac{\pi}{4} \), quadrant I: \( \sqrt{\dfrac{1 - \frac{\sqrt2}{2}}{2}} = \dfrac{\sqrt{2 - \sqrt2}}{2} \approx 0.3827 \). \( \dfrac{\sqrt{2 - \sqrt2}}{2} \)

8. If \( \cos\theta = -\dfrac{5}{13} \) with \( \theta \) in quadrant III, find \( \sin\dfrac{\theta}{2} \).
   Show the full solution

   \( \theta \) between \( \pi \) and \( \dfrac{3\pi}{2} \), so \( \dfrac{\theta}{2} \) is between \( \dfrac{\pi}{2} \) and \( \dfrac{3\pi}{4} \): quadrant II, where sine is positive. \( \sqrt{\dfrac{1 + 5/13}{2}} = \sqrt{\dfrac{18/13}{2}} = \sqrt{\dfrac{9}{13}} = \dfrac{3}{\sqrt{13}} \approx 0.832 \). \( \dfrac{3\sqrt{13}}{13} \)

9. Prove \( \dfrac{\sin 2x}{1 + \cos 2x} = \tan x \).
   Show the full solution

   \( \dfrac{2\sin x\cos x}{1 + (2\cos^2 x - 1)} = \dfrac{2\sin x\cos x}{2\cos^2 x} = \tan x \). ∎ Choose the form \( 2\cos^2 x - 1 \)

10. Show that the three forms of \( \cos 2x \) agree at \( x = 30^\circ \), and explain how they are related.
    Show the full solution

    \( \cos^2 - \sin^2 = \dfrac34 - \dfrac14 = \dfrac12 \). \( 2\cos^2 - 1 = \dfrac32 - 1 = \dfrac12 \). \( 1 - 2\sin^2 = 1 - \dfrac12 = \dfrac12 \). And \( \cos 60^\circ = \dfrac12 \). The second and third come from the first by replacing \( \sin^2 x \) with \( 1 - \cos^2 x \) or \( \cos^2 x \) with \( 1 - \sin^2 x \), using the Pythagorean identity. All equal \( \dfrac12 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \sin x = \dfrac12 \) on \( [0, 2\pi) \).
   Show the full solution

   Quadrants I and II with reference angle \( \dfrac{\pi}{6} \). \( \dfrac{\pi}{6}, \dfrac{5\pi}{6} \)

2. Solve \( \cos x = \dfrac12 \) on \( [0, 2\pi) \).
   Show the full solution

   Quadrants I and IV with reference angle \( \dfrac{\pi}{3} \). \( \dfrac{\pi}{3}, \dfrac{5\pi}{3} \)

3. Solve \( \tan x = 1 \) on \( [0, 2\pi) \).
   Show the full solution

   Quadrants I and III, reference angle \( \dfrac{\pi}{4} \). \( \dfrac{\pi}{4}, \dfrac{5\pi}{4} \)

4. Solve \( \sin x = -1 \) on \( [0, 2\pi) \).
   Show the full solution

   The bottom of the circle. \( \dfrac{3\pi}{2} \)

5. Solve \( \cos x = 2 \).
   Show the full solution

   Cosine never exceeds 1. No solution

6. Solve \( 2\sin x + \sqrt3 = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \sin x = -\dfrac{\sqrt3}{2} \): quadrants III and IV, reference \( \dfrac{\pi}{3} \). \( x = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3} \) and \( x = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3} \). \( \dfrac{4\pi}{3}, \dfrac{5\pi}{3} \)

7. Solve \( \cos x = -0.4 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \cos^{-1}(-0.4) = 1.9823 \), in quadrant II. By symmetry the other is \( 2\pi - 1.9823 = 4.3009 \), in quadrant III. Check: \( \cos 4.3009 = -0.4 \). 1.9823 and 4.3009

8. Give the general solution of \( \tan x = \sqrt3 \).
   Show the full solution

   One solution is \( \dfrac{\pi}{3} \), and tangent repeats every \( \pi \). \( x = \dfrac{\pi}{3} + k\pi \), for integer \( k \)

9. Solve \( 3\tan x = -3 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \tan x = -1 \): quadrants II and IV. \( \dfrac{3\pi}{4}, \dfrac{7\pi}{4} \)

10. Solve \( \sin x = 0.3 \) for all \( x \), and explain why the calculator gives one value.
    Show the full solution

    \( \sin^{-1}0.3 = 0.3047 \). The second solution is \( \pi - 0.3047 = 2.8369 \). General: \( x = 0.3047 + 2k\pi \) or \( x = 2.8369 + 2k\pi \). The calculator's inverse function returns only the value in \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \); the others are found by symmetry and periodicity. Two families of solutions, one from the calculator

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \sin x(2\cos x - 1) = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \sin x = 0 \): \( 0, \pi \). \( \cos x = \dfrac12 \): \( \dfrac{\pi}{3}, \dfrac{5\pi}{3} \). \( 0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3} \)

2. Solve \( 2\sin^2 x - \sin x = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \sin x(2\sin x - 1) = 0 \): \( \sin x = 0 \) or \( \dfrac12 \). \( 0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi \)

3. Solve \( \tan^2 x = 3 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \tan x = \pm\sqrt3 \): reference angle \( \dfrac{\pi}{3} \) in all four quadrants. \( \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3} \)

4. Solve \( 2\cos^2 x + \cos x - 1 = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( (2\cos x - 1)(\cos x + 1) = 0 \): \( \cos x = \dfrac12 \) or \( -1 \). \( \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3} \)

5. Solve \( \sin x = \cos x \) on \( [0, 2\pi) \).
   Show the full solution

   Divide by \( \cos x \), which is safe here because \( \cos x = 0 \) would make \( \sin x = \pm 1 \), not equal: \( \tan x = 1 \). \( \dfrac{\pi}{4}, \dfrac{5\pi}{4} \)

6. Solve \( 2\sin^2 x + 3\sin x + 1 = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( (2\sin x + 1)(\sin x + 1) = 0 \): \( \sin x = -\dfrac12 \) or \( -1 \). \( \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6} \)

7. Solve \( \cos 2x = \cos x \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2\cos^2 x - \cos x - 1 = 0 \), so \( (2\cos x + 1)(\cos x - 1) = 0 \). \( 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \)

8. Solve \( 2\cos^2 x + 3\sin x - 3 = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   Use \( \cos^2 x = 1 - \sin^2 x \): \( 2\sin^2 x - 3\sin x + 1 = 0 \), so \( (2\sin x - 1)(\sin x - 1) = 0 \). \( \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6} \)

9. Solve \( \sin 2x = \sin x \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2\sin x\cos x - \sin x = 0 \), so \( \sin x(2\cos x - 1) = 0 \). \( \sin x = 0 \): \( 0, \pi \). \( \cos x = \dfrac12 \): \( \dfrac{\pi}{3}, \dfrac{5\pi}{3} \). Check \( x = \pi \): \( \sin 2\pi = 0 = \sin\pi \) ✓. \( 0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3} \)

10. A student solves \( \sin x\cos x = \sin x \) by dividing by \( \sin x \), getting \( \cos x = 1 \) and \( x = 0 \). What is missing?
    Show the full solution

    Dividing by \( \sin x \) assumes it is not zero, and it is zero at \( x = 0 \) and \( x = \pi \). Factoring gives \( \sin x(\cos x - 1) = 0 \): the solutions are \( 0 \) and \( \pi \). The student found \( 0 \) only by accident (it satisfies both factors) and missed \( x = \pi \). Check \( x = \pi \): \( 0 \cdot (-1) = 0 = \sin\pi \) ✓. \( x = \pi \) is missing; factor, never divide

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \sin 2x = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2x = 0, \pi, 2\pi, 3\pi \) on \( [0, 4\pi) \); divide by 2. \( 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} \)

2. Solve \( \cos 2x = \dfrac12 \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}, \dfrac{7\pi}{3}, \dfrac{11\pi}{3} \). \( \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \)

3. Solve \( \tan 2x = 1 \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2x = \dfrac{\pi}{4} + k\pi \), so \( x = \dfrac{\pi}{8} + \dfrac{k\pi}{2} \). \( \dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8} \)

4. Solve \( \sin\dfrac{x}{2} = \dfrac12 \) on \( [0, 2\pi) \).
   Show the full solution

   \( \dfrac{x}{2} \in [0, \pi) \): \( \dfrac{x}{2} = \dfrac{\pi}{6}, \dfrac{5\pi}{6} \). \( \dfrac{\pi}{3}, \dfrac{5\pi}{3} \)

5. Solve \( \cos 3x = 1 \) on \( [0, 2\pi) \).
   Show the full solution

   \( 3x = 0, 2\pi, 4\pi \) on \( [0, 6\pi) \). \( 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \)

6. Solve \( 2\sin 3x = -1 \) on \( [0, 2\pi) \).
   Show the full solution

   Six solutions, as worked in the example. \( \dfrac{7\pi}{18}, \dfrac{11\pi}{18}, \dfrac{19\pi}{18}, \dfrac{23\pi}{18}, \dfrac{31\pi}{18}, \dfrac{35\pi}{18} \)

7. Solve \( \sin 2x = \cos x \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2\sin x\cos x - \cos x = 0 \), so \( \cos x(2\sin x - 1) = 0 \). \( \cos x = 0 \): \( \dfrac{\pi}{2}, \dfrac{3\pi}{2} \). \( \sin x = \dfrac12 \): \( \dfrac{\pi}{6}, \dfrac{5\pi}{6} \). \( \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2} \)

8. Solve \( \cos 2x + 3\cos x + 2 = 0 \) on \( [0, 2\pi) \).
   Show the full solution

   \( 2\cos^2 x - 1 + 3\cos x + 2 = 0 \), so \( 2\cos^2 x + 3\cos x + 1 = 0 \) and \( (2\cos x + 1)(\cos x + 1) = 0 \). \( \cos x = -\dfrac12 \): \( \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \). \( \cos x = -1 \): \( \pi \). \( \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3} \)

9. Give all real solutions of \( \sin 2x = \dfrac{\sqrt3}{2} \).
   Show the full solution

   \( 2x = \dfrac{\pi}{3} + 2k\pi \) or \( \dfrac{2\pi}{3} + 2k\pi \); divide by 2. \( x = \dfrac{\pi}{6} + k\pi \) or \( x = \dfrac{\pi}{3} + k\pi \)

10. How many solutions does \( \sin 4x = 0.5 \) have on \( [0, 2\pi) \)? List them.
    Show the full solution

    \( u = 4x \) runs over \( [0, 8\pi) \), which is four turns, with two solutions per turn: eight solutions. \( u = \dfrac{\pi}{6} + 2k\pi \) and \( \dfrac{5\pi}{6} + 2k\pi \) for \( k = 0, 1, 2, 3 \), so \( x = \dfrac{\pi}{24} + \dfrac{k\pi}{2} \) and \( \dfrac{5\pi}{24} + \dfrac{k\pi}{2} \): \( \dfrac{\pi}{24}, \dfrac{5\pi}{24}, \dfrac{13\pi}{24}, \dfrac{17\pi}{24}, \dfrac{25\pi}{24}, \dfrac{29\pi}{24}, \dfrac{37\pi}{24}, \dfrac{41\pi}{24} \). Check the last: \( 4x = \dfrac{41\pi}{6} = 6\pi + \dfrac{5\pi}{6} \) and \( \sin\dfrac{5\pi}{6} = 0.5 \). Eight solutions

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. The legs of a right triangle are 3 and 4. Find the hypotenuse and the acute angles.
   Show the full solution

   Hypotenuse \( \sqrt{9 + 16} = 5 \). \( \tan A = \dfrac34 \), so \( A = 36.87^\circ \) and \( B = 53.13^\circ \). 5; \( 36.87^\circ \) and \( 53.13^\circ \)

2. A right triangle has hypotenuse 10 and an angle of \( 40^\circ \). Find both legs.
   Show the full solution

   Opposite \( 10\sin 40^\circ = 6.428 \); adjacent \( 10\cos 40^\circ = 7.660 \). Check: \( 41.32 + 58.68 = 100 \). 6.43 and 7.66

3. The angle of elevation to a building top from 100 m away is \( 30^\circ \). Find its height.
   Show the full solution

   \( h = 100\tan 30^\circ = 100(0.5774) \). 57.74 m

4. A kite string 50 m long makes \( 55^\circ \) with the ground. Find the kite's height.
   Show the full solution

   \( 50\sin 55^\circ = 50(0.81915) \). 40.96 m

5. Find the acute angle with \( \tan\theta = 0.75 \).
   Show the full solution

   \( \theta = \tan^{-1}(0.75) \). \( 36.87^\circ \)

6. From a tower 80 m high the angle of depression to a car is \( 20^\circ \). How far is the car from the base?
   Show the full solution

   \( d = \dfrac{80}{\tan 20^\circ} = \dfrac{80}{0.36397} = 219.8 \). About 219.8 m

7. A ramp rises 1.5 m over a horizontal distance of 18 m. Find the angle of the ramp and its length.
   Show the full solution

   \( \tan\theta = \dfrac{1.5}{18} \), so \( \theta = 4.76^\circ \). Length: \( \sqrt{18^2 + 1.5^2} = \sqrt{326.25} = 18.06 \). \( 4.76^\circ \); 18.06 m

8. From one point the elevation to a tower top is \( 30^\circ \); 40 m closer it is \( 45^\circ \). Find the tower's height.
   Show the full solution

   \( h(\sqrt3 - 1) = 40 \), so \( h = \dfrac{40}{\sqrt3 - 1} = 54.64 \). Check: \( 54.64\sqrt3 - 54.64 = 94.64 - 54.64 = 40 \). 54.64 m

9. Two observers 500 m apart stand on opposite sides of a tall balloon, in line with the point beneath it. Their angles of elevation to the balloon are \( 50^\circ \) and \( 40^\circ \). Find the balloon's height, and say how this differs from the two-position problem in the worked example.
   Show the full solution

   With the base between the observers, the two distances add to 500: \( \dfrac{h}{\tan 50^\circ} + \dfrac{h}{\tan 40^\circ} = 500 \). \( h(0.83910 + 1.19175) = 500 \), so \( h = \dfrac{500}{2.03085} = 246.2 \). Check: \( 246.2/\tan 50^\circ = 206.6 \) and \( 246.2/\tan 40^\circ = 293.4 \); sum 500. The structure is the same as before, two tangents with one shared height, but the distances add instead of subtract. 246.2 m

10. An isosceles triangle has sides 10, 10 and 12. Find its height and its angles by cutting it into right triangles.
    Show the full solution

    The altitude to the base of 12 bisects it: a right triangle with hypotenuse 10 and base 6. Height \( \sqrt{100 - 36} = 8 \). Base angle \( \cos^{-1}\dfrac{6}{10} = 53.13^\circ \) (twice, since the triangle is isosceles). Apex angle \( 180^\circ - 2(53.13^\circ) = 73.74^\circ \). Check: \( 2\sin^{-1}\dfrac{6}{10} = 2(36.87^\circ) = 73.74^\circ \). Height 8; angles \( 53.13^\circ, 53.13^\circ, 73.74^\circ \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. In a triangle \( a = 8 \), \( A = 30^\circ \), \( B = 60^\circ \). Find \( b \).
   Show the full solution

   \( b = \dfrac{8\sin 60^\circ}{\sin 30^\circ} = \dfrac{8(0.8660)}{0.5} \). 13.86

2. Find the third angle if \( A = 50^\circ \) and \( B = 60^\circ \).
   Show the full solution

   \( 180^\circ - 110^\circ \). \( 70^\circ \)

3. Find \( b \) if \( a = 5 \), \( A = 40^\circ \), \( B = 70^\circ \).
   Show the full solution

   \( \dfrac{5\sin 70^\circ}{\sin 40^\circ} = \dfrac{5(0.93969)}{0.64279} \). 7.31

4. Find \( c \) if \( A = 35^\circ \), \( B = 75^\circ \), \( a = 10 \).
   Show the full solution

   \( C = 70^\circ \). \( c = \dfrac{10\sin 70^\circ}{\sin 35^\circ} = \dfrac{9.3969}{0.57358} \). 16.38

5. Find angle \( B \) if \( a = 9 \), \( b = 7 \), \( A = 80^\circ \).
   Show the full solution

   \( \sin B = \dfrac{7\sin 80^\circ}{9} = \dfrac{6.894}{9} = 0.7659 \), so \( B = 50.0^\circ \). The obtuse alternative \( 130^\circ \) would make the angle sum exceed \( 180^\circ \) with \( A = 80^\circ \), so it is excluded. \( 50.0^\circ \)

6. A triangle has angles \( 40^\circ, 60^\circ, 80^\circ \) and shortest side 10. Find the other sides.
   Show the full solution

   The shortest side faces \( 40^\circ \). The side opposite \( 60^\circ \): \( \dfrac{10\sin 60^\circ}{\sin 40^\circ} = 13.47 \). Opposite \( 80^\circ \): \( \dfrac{10\sin 80^\circ}{\sin 40^\circ} = 15.32 \). 13.47 and 15.32

7. Find \( a \) if \( b = 14 \), \( A = 25^\circ \), \( B = 100^\circ \).
   Show the full solution

   \( a = \dfrac{14\sin 25^\circ}{\sin 100^\circ} = \dfrac{14(0.42262)}{0.98481} = 6.008 \). The obtuse angle \( 100^\circ \) is opposite the longest side, \( b \), as expected. 6.01

8. Two observers 100 m apart sight a tree across a river at \( 50^\circ \) and \( 70^\circ \). Find the distance from the observer at the \( 50^\circ \) end to the tree.
   Show the full solution

   The angle at the tree is \( 60^\circ \). The distance from the \( 50^\circ \) observer is opposite the \( 70^\circ \) angle: \( \dfrac{100\sin 70^\circ}{\sin 60^\circ} = 108.5 \). 108.5 m

9. Prove the law of sines using an altitude, and check it numerically for the triangle \( A = 40^\circ \), \( B = 65^\circ \), \( a = 12 \), \( b = 16.92 \).
   Show the full solution

   The altitude gives \( h = b\sin A = a\sin B \). Then \( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} \). Numerically: \( 16.92\sin 40^\circ = 10.88 \) and \( 12\sin 65^\circ = 10.88 \); the ratios are \( \dfrac{12}{0.64279} = 18.67 \) and \( \dfrac{16.92}{0.90631} = 18.67 \). Both ratios equal 18.67

10. Explain why the law of sines cannot solve a triangle when only the three sides are given.
    Show the full solution

    Each ratio in the law pairs a side with the sine of its opposite angle, and with three sides and no angle there is no complete pair to start from. Every proportion has at least two unknowns. The law of cosines, which relates three sides to one angle, is needed first, and the law of sines can be used afterward. No side-angle pair is known

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( h \) for \( A = 30^\circ \), \( b = 10 \), and say how many triangles exist if \( a = 5 \).
   Show the full solution

   \( h = 10\sin 30^\circ = 5 = a \). One right triangle

2. How many triangles have \( A = 30^\circ \), \( b = 10 \), \( a = 3 \)?
   Show the full solution

   \( h = 5 \gt 3 \). None

3. How many have \( A = 30^\circ \), \( b = 10 \), \( a = 8 \)?
   Show the full solution

   \( h = 5 \lt 8 \lt 10 \). Two

4. How many have \( A = 30^\circ \), \( b = 10 \), \( a = 12 \)?
   Show the full solution

   \( a \gt b \), acute \( A \). One

5. For \( A = 30^\circ \), \( a = 12 \), \( b = 20 \), find both possible values of \( B \).
   Show the full solution

   \( \sin B = 0.8333 \): \( 56.44^\circ \) and \( 123.56^\circ \). \( 56.44^\circ \) and \( 123.56^\circ \)

6. Solve the triangle \( A = 30^\circ \), \( a = 25 \), \( b = 20 \).
   Show the full solution

   \( \sin B = 0.4 \), \( B = 23.58^\circ \), \( C = 126.42^\circ \), \( c = \dfrac{25\sin 126.42^\circ}{0.5} = 40.23 \). \( B = 23.58^\circ \), \( C = 126.42^\circ \), \( c = 40.23 \)

7. For \( A = 40^\circ \), \( a = 10 \), \( b = 12 \), decide the number of triangles and find the possible values of \( B \).
   Show the full solution

   \( h = 12\sin 40^\circ = 7.71 \lt 10 \lt 12 \): two triangles. \( \sin B = \dfrac{12\sin 40^\circ}{10} = 0.7713 \): \( B = 50.47^\circ \) or \( 129.53^\circ \). Both fit: \( 40^\circ + 129.53^\circ = 169.53^\circ \lt 180^\circ \). Two; \( 50.47^\circ \) and \( 129.53^\circ \)

8. How many triangles have \( A = 50^\circ \), \( b = 8 \), \( a = 5 \)?
   Show the full solution

   \( h = 8\sin 50^\circ = 6.13 \gt 5 \). None

9. Solve \( A = 120^\circ \), \( a = 10 \), \( b = 6 \), and explain why there is one triangle.
   Show the full solution

   \( \sin B = \dfrac{6\sin 120^\circ}{10} = 0.5196 \), so \( B = 31.31^\circ \). The alternative \( 148.69^\circ \) with \( A = 120^\circ \) exceeds \( 180^\circ \). With an obtuse \( A \), no second angle can also be obtuse, and \( a \gt b \) is needed for any triangle at all. One triangle, \( B = 31.31^\circ \)

10. Explain geometrically why two triangles can arise in the case \( h \lt a \lt b \).
    Show the full solution

    Fix the angle \( A \) and the side \( b \) that runs along one ray of it. The far end of \( b \) is a point at distance \( h = b\sin A \) from the other ray. Side \( a \) is a segment from that point to the other ray, so swing it like a compass arm of length \( a \). When \( a \) is longer than \( h \) but shorter than \( b \), the arc crosses the ray at two points, one on each side of the foot of the perpendicular. The two crossing points give angles \( B \) and \( 180^\circ - B \), which share the same sine, and both fit inside a triangle with \( A \). The arc meets the base line twice

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( c \) if \( a = 5 \), \( b = 7 \), \( C = 90^\circ \).
   Show the full solution

   The cosine is 0, so \( c^2 = 25 + 49 = 74 \). \( c = 8.60 \)

2. Find \( c \) if \( a = 6 \), \( b = 8 \), \( C = 60^\circ \).
   Show the full solution

   \( c^2 = 36 + 64 - 2(6)(8)(0.5) = 100 - 48 = 52 \). \( c = 7.21 \)

3. Find \( c \) if \( a = b = 10 \) and \( C = 36^\circ \).
   Show the full solution

   \( c^2 = 200 - 200\cos 36^\circ = 200 - 161.80 = 38.20 \). \( c = 6.18 \)

4. A triangle has sides 3, 4, 5. Find the largest angle.
   Show the full solution

   \( \cos C = \dfrac{9 + 16 - 25}{24} = 0 \). \( 90^\circ \)

5. What does \( \cos C = 0 \) tell you about the triangle?
   Show the full solution

   The angle is \( 90^\circ \), and the law reduces to the Pythagorean theorem. It is a right angle

6. Find the largest angle of the triangle with sides 5, 6, 7.
   Show the full solution

   \( \cos C = \dfrac{25 + 36 - 49}{2(5)(6)} = \dfrac{12}{60} = 0.2 \), so \( C = 78.46^\circ \). \( 78.46^\circ \)

7. Two ships leave a port on paths \( 120^\circ \) apart at 20 km/h and 30 km/h. How far apart are they after 2 hours?
   Show the full solution

   Sides 40 and 60 with included angle \( 120^\circ \): \( d^2 = 1600 + 3600 - 2(40)(60)\cos 120^\circ = 5200 + 2400 = 7600 \). \( d = 87.18 \). The cosine of \( 120^\circ \) is negative, so the correction adds. 87.18 km

8. Find \( c \) if \( a = 9 \), \( b = 5 \), \( C = 120^\circ \).
   Show the full solution

   \( c^2 = 81 + 25 - 2(9)(5)(-0.5) = 106 + 45 = 151 \). \( c = 12.29 \)

9. In that triangle, find angle \( A \).
   Show the full solution

   \( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{25 + 151 - 81}{2(5)(12.288)} = \dfrac{95}{122.88} = 0.7731 \), so \( A = 39.37^\circ \). Check: the angle \( 120^\circ \) faces the longest side, and \( 39.37^\circ \) is the angle opposite side 9. \( 39.37^\circ \)

10. Derive the law of cosines from coordinates.
    Show the full solution

    Put \( C = (0, 0) \), \( B = (a, 0) \), \( A = (b\cos C, b\sin C) \). Then \( c^2 = (b\cos C - a)^2 + (b\sin C)^2 = b^2 - 2ab\cos C + a^2 \), using \( \cos^2 + \sin^2 = 1 \). ∎ Check with the triangle \( a = 8 \), \( b = 11 \), \( C = 60^\circ \): \( A = (5.5, 9.526) \), \( B = (8, 0) \), distance \( \sqrt{2.5^2 + 9.526^2} = \sqrt{97} \) ✓. \( c^2 = a^2 + b^2 - 2ab\cos C \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the area with \( a = 6 \), \( b = 10 \), \( C = 30^\circ \).
   Show the full solution

   \( \tfrac12(6)(10)\sin 30^\circ = 30(0.5) \). 15

2. Find the area with \( a = 5 \), \( b = 12 \), \( C = 90^\circ \).
   Show the full solution

   \( \tfrac12(5)(12)(1) \). 30

3. Use Heron's formula for sides 3, 4, 5.
   Show the full solution

   \( s = 6 \): \( \sqrt{6 \cdot 3 \cdot 2 \cdot 1} = \sqrt{36} \). 6

4. Use Heron's formula for sides 5, 5, 6.
   Show the full solution

   \( s = 8 \): \( \sqrt{8 \cdot 3 \cdot 3 \cdot 2} = \sqrt{144} \). 12

5. Find the area of an equilateral triangle of side 6.
   Show the full solution

   \( \tfrac12(6)(6)\sin 60^\circ = 18\left(\dfrac{\sqrt3}{2}\right) = 9\sqrt3 \). \( 9\sqrt3 \approx 15.59 \)

6. Find the area with \( a = 10 \), \( b = 14 \), \( C = 40^\circ \).
   Show the full solution

   \( \tfrac12(140)\sin 40^\circ = 70(0.64279) \). 44.99

7. Find the area of the triangle with sides 13, 14, 15.
   Show the full solution

   \( s = 21 \): \( \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84 \). 84

8. A parallelogram has sides 8 and 12 and an angle of \( 50^\circ \). Find its area.
   Show the full solution

   \( ab\sin\theta = 96(0.76604) = 73.54 \). Each of the two triangles formed by a diagonal has half of this. 73.54

9. A triangular plot has sides 100 m, 120 m and 150 m. Find its area.
   Show the full solution

   \( s = 185 \): \( \sqrt{185 \cdot 85 \cdot 65 \cdot 35} = \sqrt{35{,}774{,}375} = 5981 \). About 5981 m\( ^2 \)

10. Explain why \( \tfrac12 ab\sin C \) is the same as half of base times height, and why it works for an obtuse \( C \).
    Show the full solution

    Take side \( a \) as the base. The perpendicular height from the opposite vertex is \( b\sin C \), from the right triangle containing side \( b \) and angle \( C \). So the area is \( \tfrac12 a(b\sin C) \). For an obtuse \( C \) the height lies outside the triangle, and the right triangle uses the supplementary angle \( 180^\circ - C \), with \( \sin(180^\circ - C) = \sin C \), so the same formula holds. For \( C = 90^\circ \) it reduces to \( \tfrac12 ab \). The height is \( b\sin C \) in every case

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write the bearing of due east.
   Show the full solution

   A quarter turn clockwise from north. 090

2. Write the bearing of due west.
   Show the full solution

   Three quarter turns clockwise. 270

3. A boat goes 10 km north then 10 km east. Find the distance and bearing from the start.
   Show the full solution

   Distance \( 10\sqrt2 = 14.14 \). The position is north-east. 14.14 km on bearing 045

4. A hiker walks 5 km east then 12 km north. Find the distance and bearing back to the start as seen from the start.
   Show the full solution

   Distance \( \sqrt{25 + 144} = 13 \). The bearing is \( \tan^{-1}\dfrac{5}{12} = 22.62^\circ \) east of north. 13 km on bearing 023

5. What is the back bearing of 060?
   Show the full solution

   Add \( 180^\circ \). 240

6. A plane flies 200 km on bearing 040, then 150 km on bearing 130. Find the distance and bearing from the start.
   Show the full solution

   Back bearing of the first leg is 220, and \( 220 - 130 = 90^\circ \), so the legs form a right angle. Distance \( \sqrt{200^2 + 150^2} = 250 \). Bearing \( 40^\circ + \tan^{-1}\dfrac{150}{200} = 40^\circ + 36.87^\circ = 76.87^\circ \). 250 km on bearing 077

7. Two coastguard stations 20 km apart see a ship; the angles at the two stations between the baseline and the ship are \( 60^\circ \) and \( 45^\circ \). Find each distance to the ship.
   Show the full solution

   Angle at the ship \( 75^\circ \). Distance from the \( 60^\circ \) station is opposite \( 45^\circ \): \( \dfrac{20\sin 45^\circ}{\sin 75^\circ} = 14.64 \). From the \( 45^\circ \) station: \( \dfrac{20\sin 60^\circ}{\sin 75^\circ} = 17.93 \). 14.64 km and 17.93 km

8. Name the tool for each: (i) sides 5, 7, 10; (ii) angles \( 50^\circ \), \( 60^\circ \) with side 8 between; (iii) sides 4, 6 and the included angle \( 35^\circ \).
   Show the full solution

   (i) SSS, law of cosines. (ii) ASA: third angle \( 70^\circ \), then law of sines. (iii) SAS: law of cosines for the third side. Cosines; sines after the third angle; cosines

9. A surveyor measures 120 m from A to B. A distant point T is at \( 75^\circ \) at A and \( 65^\circ \) at B from the baseline. Find AT and BT.
   Show the full solution

   Angle at T \( = 40^\circ \). \( AT \) is opposite the \( 65^\circ \) angle: \( \dfrac{120\sin 65^\circ}{\sin 40^\circ} = 169.2 \). \( BT \) is opposite the \( 75^\circ \) angle: \( \dfrac{120\sin 75^\circ}{\sin 40^\circ} = 180.3 \). 169.2 m and 180.3 m

10. A triangle has sides 5 and 7 with the \( 40^\circ \) angle opposite the side 5. Solve both possible triangles and explain why there are two.
    Show the full solution

    \( h = 7\sin 40^\circ = 4.50 \lt 5 \lt 7 \) gives two. \( \sin B = 0.8999 \): \( B = 64.15^\circ \) or \( 115.85^\circ \); \( C = 75.85^\circ \) or \( 24.15^\circ \); \( c = 7.54 \) or \( 3.18 \). Check the second with the law of sines: \( \dfrac{5}{\sin 40^\circ} = 7.78 \) and \( \dfrac{3.18}{\sin 24.15^\circ} = 7.78 \) ✓. The side of length 5 can reach the base line from the vertex in two places, since it is longer than the height 4.50 but shorter than the side 7. Triangle 1: \( B = 64.15^\circ \), \( c = 7.54 \); triangle 2: \( B = 115.85^\circ \), \( c = 3.18 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the vector from \( (2, 3) \) to \( (7, 9) \).
   Show the full solution

   \( \langle 7 - 2,\ 9 - 3 \rangle \). \( \langle 5, 6 \rangle \)

2. Find the magnitude of \( \langle 3, 4 \rangle \).
   Show the full solution

   \( \sqrt{9 + 16} \). 5

3. Find the vector from \( (-1, 4) \) to \( (3, -2) \).
   Show the full solution

   \( \langle 3 - (-1),\ -2 - 4 \rangle \). \( \langle 4, -6 \rangle \)

4. Find the magnitude of \( \langle -5, 12 \rangle \).
   Show the full solution

   \( \sqrt{25 + 144} \). 13

5. State whether speed and velocity are scalars or vectors.
   Show the full solution

   Speed has only a size; velocity also has a direction. Speed is a scalar; velocity is a vector

6. Find the components of a vector of magnitude 10 at \( 60^\circ \).
   Show the full solution

   \( \langle 10\cos 60^\circ,\ 10\sin 60^\circ \rangle = \langle 5,\ 8.660 \rangle \). Check: \( 25 + 75 = 100 \). \( \langle 5, 8.66 \rangle \)

7. Find the components of a vector of magnitude 20 at \( 135^\circ \).
   Show the full solution

   \( \cos 135^\circ = -\dfrac{\sqrt2}{2} \) and \( \sin 135^\circ = \dfrac{\sqrt2}{2} \). \( \langle -14.14,\ 14.14 \rangle \). Quadrant II: negative horizontal, positive vertical. \( \langle -10\sqrt2, 10\sqrt2 \rangle \)

8. A vector \( \langle 3, -2 \rangle \) starts at \( (1, 4) \). Find where it ends.
   Show the full solution

   \( (1 + 3,\ 4 - 2) \). \( (4, 2) \)

9. Are the vector from \( (0, 0) \) to \( (2, 1) \) and the vector from \( (3, 3) \) to \( (5, 4) \) equal?
   Show the full solution

   Both are \( \langle 2, 1 \rangle \). Equal vectors need equal components, not equal positions. Yes

10. Explain why a magnitude is never negative and is zero only for the zero vector, and find \( |\langle -6, 8 \rangle| \).
    Show the full solution

    The magnitude is \( \sqrt{a^2 + b^2} \), a square root of a sum of squares, so it is nonnegative; it is 0 only when both \( a^2 \) and \( b^2 \) are 0, that is \( a = b = 0 \). \( |\langle -6, 8 \rangle| = \sqrt{36 + 64} = 10 \). Direction is carried by the signs of the components, which the squares discard. 10; zero only for \( \langle 0, 0 \rangle \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \langle 1, 2 \rangle + \langle 3, 4 \rangle \).
   Show the full solution

   \( \langle 4, 6 \rangle \)

2. Find \( 3\langle 2, -1 \rangle \).
   Show the full solution

   \( \langle 6, -3 \rangle \)

3. Find \( \langle 5, 3 \rangle - \langle 2, 7 \rangle \).
   Show the full solution

   \( \langle 3, -4 \rangle \)

4. Find \( 2\langle 1, 4 \rangle + \langle -3, 1 \rangle \).
   Show the full solution

   \( \langle 2, 8 \rangle + \langle -3, 1 \rangle \). \( \langle -1, 9 \rangle \)

5. Find \( -\langle 2, -5 \rangle \).
   Show the full solution

   Scale by \( -1 \). \( \langle -2, 5 \rangle \)

6. If \( \vec u = \langle 4, 1 \rangle \) and \( \vec v = \langle -1, 3 \rangle \), find \( 3\vec u - 2\vec v \).
   Show the full solution

   \( \langle 12, 3 \rangle - \langle -2, 6 \rangle = \langle 14, -3 \rangle \). \( \langle 14, -3 \rangle \)

7. Solve for \( \vec x \): \( 2\vec x + \langle 1, 2 \rangle = \langle 7, -4 \rangle \).
   Show the full solution

   \( 2\vec x = \langle 6, -6 \rangle \), so \( \vec x = \langle 3, -3 \rangle \). Check: \( 2\langle 3, -3 \rangle + \langle 1, 2 \rangle = \langle 7, -4 \rangle \) ✓. \( \langle 3, -3 \rangle \)

8. Are \( \langle 2, 3 \rangle \) and \( \langle -6, -9 \rangle \) parallel?
   Show the full solution

   \( \langle -6, -9 \rangle = -3\langle 2, 3 \rangle \). Yes, pointing in opposite directions

9. For \( \vec u = \langle 3, 4 \rangle \) and \( \vec v = \langle 5, 12 \rangle \), compare \( |\vec u + \vec v| \) with \( |\vec u| + |\vec v| \).
   Show the full solution

   \( \vec u + \vec v = \langle 8, 16 \rangle \), magnitude \( \sqrt{320} = 17.89 \). \( |\vec u| + |\vec v| = 5 + 13 = 18 \). \( 17.89 \lt 18 \)

10. When is \( |\vec u + \vec v| = |\vec u| + |\vec v| \)? Illustrate with \( \langle 3, 4 \rangle \) and \( \langle 6, 8 \rangle \).
    Show the full solution

    Equality holds when the vectors are parallel and point the same way, so no length is lost to cancellation. \( \langle 3, 4 \rangle + \langle 6, 8 \rangle = \langle 9, 12 \rangle \), magnitude 15, and \( 5 + 10 = 15 \). For antiparallel vectors the magnitude of the sum is the difference of the magnitudes. Parallel, same direction

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the magnitude of \( \langle 6, 8 \rangle \).
   Show the full solution

   10

2. Find the unit vector in the direction of \( \langle 3, 4 \rangle \).
   Show the full solution

   Divide by 5. \( \left\langle \dfrac35, \dfrac45 \right\rangle \)

3. Find the direction angle of \( \langle 1, 1 \rangle \).
   Show the full solution

   \( \tan\theta = 1 \), quadrant I. \( 45^\circ \)

4. Write \( \langle 2, -7 \rangle \) using \( \vec i \) and \( \vec j \).
   Show the full solution

   \( 2\vec i - 7\vec j \)

5. Is \( \langle 0.6, 0.8 \rangle \) a unit vector?
   Show the full solution

   \( 0.36 + 0.64 = 1 \). Yes

6. Find the direction angle of \( \langle -1, \sqrt3 \rangle \).
   Show the full solution

   Reference angle \( \tan^{-1}\sqrt3 = 60^\circ \); quadrant II gives \( 180^\circ - 60^\circ \). \( 120^\circ \)

7. Find the direction angle of \( \langle -2, -2 \rangle \).
   Show the full solution

   Reference \( 45^\circ \), quadrant III. \( 225^\circ \)

8. Find the unit vector in the direction of \( \langle -5, 12 \rangle \).
   Show the full solution

   Magnitude 13. \( \left\langle -\dfrac{5}{13}, \dfrac{12}{13} \right\rangle \)

9. Find the vector of magnitude 26 in the direction of \( \langle 5, -12 \rangle \).
   Show the full solution

   \( 26 \times \left\langle \dfrac{5}{13}, -\dfrac{12}{13} \right\rangle \). \( \langle 10, -24 \rangle \)

10. Explain why \( \tan^{-1}\dfrac{b}{a} \) fails for \( \langle -3, -4 \rangle \), and give the correct direction.
    Show the full solution

    The ratio \( \dfrac{-4}{-3} = \dfrac43 \) hides the signs of both components, and the inverse tangent returns \( 53.13^\circ \) in quadrant I. The vector is in quadrant III, so add \( 180^\circ \): \( 233.13^\circ \). Check: \( 5\cos 233.13^\circ = -3 \) and \( 5\sin 233.13^\circ = -4 \). \( 233.13^\circ \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the resultant of \( \langle 3, 4 \rangle \) and \( \langle 2, -1 \rangle \).
   Show the full solution

   \( \langle 5, 3 \rangle \)

2. Find the magnitude of \( \langle 5, 3 \rangle \).
   Show the full solution

   \( \sqrt{25 + 9} = \sqrt{34} \). 5.83

3. Two forces of 10 N, one east and one north, act on an object. Find the resultant.
   Show the full solution

   \( \langle 10, 10 \rangle \): magnitude \( 10\sqrt2 \) at \( 45^\circ \). 14.14 N at \( 45^\circ \) north of east

4. A plane flies east at 300 km/h in a north wind of 50 km/h (blowing south). Find the ground speed.
   Show the full solution

   \( \langle 300, -50 \rangle \): \( \sqrt{90000 + 2500} = 304.1 \). 304.1 km/h

5. Three forces \( \langle 3, -2 \rangle \), \( \langle -5, 4 \rangle \) and \( \vec F \) balance. Find \( \vec F \).
   Show the full solution

   The sum of all three is zero: \( \vec F = -(\langle 3, -2 \rangle + \langle -5, 4 \rangle) = -\langle -2, 2 \rangle \). \( \langle 2, -2 \rangle \)

6. A 100 N weight hangs from two cables, each at \( 30^\circ \) above the horizontal on opposite sides. Find the tension in each.
   Show the full solution

   Symmetry makes the tensions equal. Vertical balance: \( 2T\sin 30^\circ = 100 \), so \( T = 100 \) N. The horizontal components cancel. 100 N each

7. A swimmer heads straight across a 60 m river at 2 m/s, with a current of 1.5 m/s. Find the crossing time, the downstream drift and the ground speed.
   Show the full solution

   The crossing speed is 2 m/s, so \( t = \dfrac{60}{2} = 30 \) s. Drift \( 1.5 \times 30 = 45 \) m. Ground speed \( \sqrt{4 + 2.25} = 2.5 \) m/s. 30 s; 45 m; 2.5 m/s

8. The swimmer wants to land directly opposite. At what angle upstream from straight across should she aim, and how long does the crossing take?
   Show the full solution

   The upstream component must cancel the current: \( 2\sin\varphi = 1.5 \), so \( \varphi = \sin^{-1}0.75 = 48.59^\circ \). The speed across is \( 2\cos 48.59^\circ = 1.323 \) m/s. \( t = \dfrac{60}{1.323} = 45.4 \) s. Check: \( \sqrt{4 - 2.25} = 1.323 \). \( 48.59^\circ \) upstream; 45.4 s

9. A 100 N weight hangs from cables at \( 30^\circ \) and \( 60^\circ \). Find the tensions and check the horizontal balance.
   Show the full solution

   \( T_1 = 50 \) N (at \( 30^\circ \)) and \( T_2 = 86.60 \) N (at \( 60^\circ \)). Horizontal: \( 50\cos 30^\circ = 43.30 = 86.60\cos 60^\circ \) ✓. Vertical: \( 25 + 75 = 100 \) ✓. 50 N and 86.60 N

10. Explain why equilibrium requires both the horizontal and the vertical component sums to be zero, not just the total magnitude.
    Show the full solution

    A vector is zero only if every component is zero, since its magnitude is \( \sqrt{R_x^2 + R_y^2} \), which vanishes only when both terms do. Adding magnitudes of forces is meaningless for this: two forces of 50 N in different directions do not cancel even though the "total" looks balanced. Each component sum can be treated as its own one-dimensional balance, which is what makes the two-equation method work. A vector is zero only if each component is zero

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \langle 2, 3 \rangle \cdot \langle 4, 1 \rangle \).
   Show the full solution

   \( 8 + 3 \). 11

2. Find \( \langle 1, 0 \rangle \cdot \langle 0, 1 \rangle \).
   Show the full solution

   \( 0 + 0 \); the axes are perpendicular. 0

3. Find the angle between \( \langle 1, 0 \rangle \) and \( \langle 1, 1 \rangle \).
   Show the full solution

   \( \cos\theta = \dfrac{1}{1 \cdot \sqrt2} \). \( 45^\circ \)

4. Are \( \langle 3, 2 \rangle \) and \( \langle -2, 3 \rangle \) orthogonal?
   Show the full solution

   \( -6 + 6 = 0 \). Yes

5. Find \( \langle 1, 2 \rangle \cdot \langle 1, 2 \rangle \).
   Show the full solution

   \( 1 + 4 = 5 = |\vec u|^2 \). 5

6. Find the angle between \( \langle 2, 1 \rangle \) and \( \langle 1, 3 \rangle \).
   Show the full solution

   Dot \( 2 + 3 = 5 \); magnitudes \( \sqrt5 \) and \( \sqrt{10} \). \( \cos\theta = \dfrac{5}{\sqrt{50}} = 0.7071 \). \( 45^\circ \)

7. Find \( k \) so that \( \langle k, 3 \rangle \) is perpendicular to \( \langle 4, -6 \rangle \).
   Show the full solution

   \( 4k - 18 = 0 \). \( k = 4.5 \)

8. Find the angle between \( \langle -1, 2 \rangle \) and \( \langle 3, 1 \rangle \).
   Show the full solution

   Dot \( -3 + 2 = -1 \). \( \cos\theta = \dfrac{-1}{\sqrt5\sqrt{10}} = -0.14142 \), so \( \theta = 98.13^\circ \). Obtuse, as the negative dot product says. \( 98.13^\circ \)

9. Verify \( |\vec u + \vec v|^2 = |\vec u|^2 + |\vec v|^2 + 2\vec u \cdot \vec v \) for \( \vec u = \langle 1, 2 \rangle \), \( \vec v = \langle 3, -1 \rangle \).
   Show the full solution

   \( \vec u + \vec v = \langle 4, 1 \rangle \), so the left side is 17. Right side: \( 5 + 10 + 2(1) = 17 \) ✓. Both 17

10. Explain what the sign of a dot product says about the angle, and classify \( \langle 2, 5 \rangle \cdot \langle -5, 2 \rangle \), \( \langle 1, 1 \rangle \cdot \langle 2, 3 \rangle \) and \( \langle 1, 0 \rangle \cdot \langle -1, 1 \rangle \).
    Show the full solution

    \( \vec u \cdot \vec v = |\vec u||\vec v|\cos\theta \), and the magnitudes are positive, so the sign is the sign of \( \cos\theta \): positive for an acute angle, zero for \( 90^\circ \), negative for an obtuse angle. \( -10 + 10 = 0 \): perpendicular. \( 2 + 3 = 5 \gt 0 \): acute. \( -1 + 0 = -1 \lt 0 \): obtuse. Perpendicular, acute, obtuse

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A force \( \langle 10, 0 \rangle \) N moves an object \( \langle 5, 0 \rangle \) m. Find the work.
   Show the full solution

   \( \vec F \cdot \vec d = 50 \). 50 J

2. A 20 N force acts along a 6 m displacement. Find the work.
   Show the full solution

   The angle is \( 0^\circ \): \( 20 \times 6 \times 1 \). 120 J

3. How much work does a force perpendicular to the motion do?
   Show the full solution

   \( \cos 90^\circ = 0 \). 0

4. Find the work of \( \vec F = \langle 3, 4 \rangle \) N over \( \vec d = \langle 2, 1 \rangle \) m.
   Show the full solution

   \( 6 + 4 \). 10 J

5. Find the scalar component of \( \langle 3, 4 \rangle \) along \( \langle 1, 0 \rangle \).
   Show the full solution

   \( \dfrac{3}{1} \). 3

6. Project \( \langle 2, 3 \rangle \) onto \( \langle 4, 1 \rangle \).
   Show the full solution

   \( \dfrac{11}{17}\langle 4, 1 \rangle \). \( \left\langle \dfrac{44}{17}, \dfrac{11}{17} \right\rangle \)

7. Find the work of a 50 N force at \( 30^\circ \) to a horizontal displacement of 8 m.
   Show the full solution

   \( 50 \times 8 \times \cos 30^\circ = 400(0.86603) \). 346.4 J

8. Find the components of a 200 N weight along and perpendicular to a \( 25^\circ \) ramp.
   Show the full solution

   Along: \( 200\sin 25^\circ = 84.52 \). Perpendicular: \( 200\cos 25^\circ = 181.26 \). Check: \( 84.52^2 + 181.26^2 = 40000 \). 84.5 N and 181.3 N

9. A 100 N force at \( 60^\circ \) to the horizontal pulls a cart 10 m horizontally. Find the work.
   Show the full solution

   \( 100 \times 10 \times \cos 60^\circ = 1000(0.5) \). 500 J

10. Show that \( \vec u - \text{proj}_{\vec v}\vec u \) is perpendicular to \( \vec v \) in general.
    Show the full solution

    Let \( p = \text{proj}_{\vec v}\vec u = \dfrac{\vec u \cdot \vec v}{|\vec v|^2}\vec v \). Then \( (\vec u - p)\cdot\vec v = \vec u \cdot \vec v - \dfrac{\vec u \cdot \vec v}{|\vec v|^2} (\vec v \cdot \vec v) = \vec u \cdot \vec v - \vec u \cdot \vec v = 0 \). ∎ Numerically, for \( \langle 2, 3 \rangle \) on \( \langle 4, 1 \rangle \): the remainder \( \langle -0.588, 2.353 \rangle \) has dot product 0 with \( \langle 4, 1 \rangle \). The remainder has zero dot product with \( \vec v \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the magnitude of \( \langle 1, 2, 2 \rangle \).
   Show the full solution

   \( \sqrt{1 + 4 + 4} \). 3

2. Find the vector from \( (1, 0, 2) \) to \( (4, 4, 2) \) and its length.
   Show the full solution

   \( \langle 3, 4, 0 \rangle \), length \( \sqrt{9 + 16} \). \( \langle 3, 4, 0 \rangle \); 5

3. Find \( \langle 1, 2, 3 \rangle + \langle 4, -1, 2 \rangle \).
   Show the full solution

   \( \langle 5, 1, 5 \rangle \)

4. Find \( \langle 1, 2, 3 \rangle \cdot \langle 4, -1, 2 \rangle \).
   Show the full solution

   \( 4 - 2 + 6 \). 8

5. Find the distance from \( (0, 0, 0) \) to \( (2, 3, 6) \).
   Show the full solution

   \( \sqrt{4 + 9 + 36} \). 7

6. Find the unit vector in the direction of \( \langle 2, 3, 6 \rangle \).
   Show the full solution

   Divide by 7. \( \left\langle \dfrac27, \dfrac37, \dfrac67 \right\rangle \)

7. Find the angle between \( \langle 1, 2, 2 \rangle \) and \( \langle 2, -1, 2 \rangle \).
   Show the full solution

   Dot 4; magnitudes 3 and 3; \( \cos\theta = \dfrac49 \). \( 63.61^\circ \)

8. Are \( \langle 1, 2, -1 \rangle \) and \( \langle 3, -1, 1 \rangle \) perpendicular?
   Show the full solution

   \( 3 - 2 - 1 = 0 \). Yes

9. Find \( k \) so that \( \langle k, 2, 1 \rangle \) is perpendicular to \( \langle 3, -1, 4 \rangle \).
   Show the full solution

   \( 3k - 2 + 4 = 0 \), so \( k = -\dfrac23 \). Check: \( 3\left(-\dfrac23\right) - 2 + 4 = 0 \). \( k = -\dfrac23 \)

10. Show that the squares of the direction cosines of \( \langle 2, 3, 6 \rangle \) sum to 1, and explain why this must always hold.
    Show the full solution

    \( \left(\dfrac27\right)^2 + \left(\dfrac37\right)^2 + \left(\dfrac67\right)^2 = \dfrac{4 + 9 + 36}{49} = 1 \). The direction cosines are the components of the unit vector \( \hat v \), and the squares of the components of a unit vector sum to its squared magnitude, which is 1. They are the components of a unit vector

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Convert \( \left(2, \dfrac{\pi}{2}\right) \) to rectangular coordinates.
   Show the full solution

   \( (2\cos\tfrac{\pi}{2}, 2\sin\tfrac{\pi}{2}) \). \( (0, 2) \)

2. Convert \( (4, 0) \) to rectangular coordinates.
   Show the full solution

   \( (4, 0) \)

3. Give a polar name for \( \left(3, \dfrac{\pi}{4}\right) \) with a negative \( r \).
   Show the full solution

   Reverse \( r \) and add \( \pi \) to the angle. \( \left(-3, \dfrac{5\pi}{4}\right) \)

4. Convert \( (5, \pi) \) to rectangular coordinates.
   Show the full solution

   \( (5\cos\pi, 5\sin\pi) \). \( (-5, 0) \)

5. Where is \( (-2, 0) \)?
   Show the full solution

   Two units from the pole, opposite the positive \( x \)-axis. \( (-2, 0) \) in rectangular coordinates

6. Convert \( \left(6, \dfrac{2\pi}{3}\right) \) to rectangular coordinates.
   Show the full solution

   \( 6\cos\dfrac{2\pi}{3} = -3 \) and \( 6\sin\dfrac{2\pi}{3} = 3\sqrt3 = 5.196 \). \( (-3, 5.196) \)

7. Convert \( \left(2, \dfrac{7\pi}{4}\right) \) to rectangular coordinates.
   Show the full solution

   \( \left(2\cdot\dfrac{\sqrt2}{2},\ 2\cdot\left(-\dfrac{\sqrt2}{2}\right)\right) \). \( (\sqrt2, -\sqrt2) \approx (1.414, -1.414) \)

8. Are \( \left(3, \dfrac{\pi}{6}\right) \) and \( \left(-3, \dfrac{7\pi}{6}\right) \) the same point?
   Show the full solution

   The second has \( r \) negated and \( \pi \) added to the angle, which names the same point. In rectangular form both are \( (2.598, 1.5) \): the second gives \( -3\cos\dfrac{7\pi}{6} = 2.598 \) and \( -3\sin\dfrac{7\pi}{6} = 1.5 \). Yes

9. Give three other polar names for \( \left(2, \dfrac{\pi}{3}\right) \).
   Show the full solution

   Add a turn, subtract a turn, or negate \( r \) and add \( \pi \). \( \left(2, \dfrac{7\pi}{3}\right) \), \( \left(2, -\dfrac{5\pi}{3}\right) \), \( \left(-2, \dfrac{4\pi}{3}\right) \)

10. Explain why a point has infinitely many polar names, and what \( (0, \theta) \) means.
    Show the full solution

    The angle can be increased by any number of full turns, giving infinitely many names with the same \( r \), and the sign of \( r \) together with a half-turn gives more. When \( r = 0 \) the point is the pole whatever the angle is, so \( (0, \theta) \) names the origin for every \( \theta \); this is the one point with no direction. Rectangular coordinates are unique because \( x \) and \( y \) do not wrap around. Angles repeat every \( 2\pi \); at the pole the angle is irrelevant

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Convert \( (1, 1) \) to polar coordinates.
   Show the full solution

   \( r = \sqrt2 \), quadrant I with equal coordinates. \( \left(\sqrt2, \dfrac{\pi}{4}\right) \)

2. Convert \( (0, 3) \) to polar coordinates.
   Show the full solution

   On the positive \( y \)-axis. \( \left(3, \dfrac{\pi}{2}\right) \)

3. Convert \( (-2, 0) \) to polar coordinates.
   Show the full solution

   On the negative \( x \)-axis. \( (2, \pi) \)

4. Convert \( (-1, -1) \) to polar coordinates.
   Show the full solution

   \( r = \sqrt2 \); quadrant III, reference \( \dfrac{\pi}{4} \). \( \left(\sqrt2, \dfrac{5\pi}{4}\right) \)

5. Convert \( r = 5 \) to rectangular form.
   Show the full solution

   Square both sides and use \( r^2 = x^2 + y^2 \). \( x^2 + y^2 = 25 \)

6. Convert \( \theta = \dfrac{\pi}{4} \) to rectangular form.
   Show the full solution

   \( \tan\theta = \dfrac{y}{x} = 1 \). \( y = x \)

7. Convert \( r = 4\cos\theta \) to rectangular form.
   Show the full solution

   \( r^2 = 4r\cos\theta \), so \( x^2 + y^2 = 4x \), and completing the square gives \( (x - 2)^2 + y^2 = 4 \). A circle of radius 2 centered at \( (2, 0) \)

8. Convert \( y = 3 \) to polar form.
   Show the full solution

   \( r\sin\theta = 3 \), so \( r = \dfrac{3}{\sin\theta} \). \( r = 3\csc\theta \)

9. Convert \( r = 2\sin\theta \) to rectangular form.
   Show the full solution

   \( r^2 = 2r\sin\theta \), so \( x^2 + y^2 = 2y \) and \( x^2 + (y - 1)^2 = 1 \). A circle of radius 1 centered at \( (0, 1) \)

10. Explain why \( \tan^{-1}\dfrac{y}{x} \) gives the wrong angle for \( (-3, 3) \) and find the right one.
    Show the full solution

    \( \dfrac{3}{-3} = -1 \), and \( \tan^{-1}(-1) = -\dfrac{\pi}{4} \), a quadrant IV angle, while the point is in quadrant II. The inverse tangent only returns angles between \( -\dfrac{\pi}{2} \) and \( \dfrac{\pi}{2} \). Add \( \pi \): \( \dfrac{3\pi}{4} \), and check \( 3\sqrt2\cos\dfrac{3\pi}{4} = -3 \). \( \dfrac{3\pi}{4} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Describe the graph of \( r = 3 \).
   Show the full solution

   Constant distance from the pole. A circle of radius 3 centered at the pole

2. Describe the graph of \( \theta = \dfrac{\pi}{3} \).
   Show the full solution

   All points at that angle, including the negative \( r \) points. A line through the pole at \( 60^\circ \)

3. Describe the graph of \( r = 2\cos\theta \).
   Show the full solution

   A circle through the pole with diameter from \( (0, 0) \) to \( (2, 0) \)

4. Find \( r \) at \( \theta = \pi \) for \( r = 1 + \cos\theta \).
   Show the full solution

   \( 1 + (-1) \). 0; the cardioid passes through the pole

5. How many petals has \( r = \sin 3\theta \)?
   Show the full solution

   \( n = 3 \) is odd. 3

6. How many petals has \( r = \cos 2\theta \)?
   Show the full solution

   \( n = 2 \) is even, so \( 2n \). 4

7. Find the maximum value of \( |r| \) for \( r = 3 + 2\sin\theta \).
   Show the full solution

   At \( \sin\theta = 1 \): \( 3 + 2 \). 5

8. Where does \( r = 1 + 2\cos\theta \) pass through the pole?
   Show the full solution

   \( \cos\theta = -\dfrac12 \) gives \( \theta = \dfrac{2\pi}{3} \) and \( \dfrac{4\pi}{3} \). At those two angles; it has an inner loop

9. For the lemniscate \( r^2 = 4\cos 2\theta \), find \( r \) at \( \theta = 0 \) and the angles where it exists.
   Show the full solution

   \( r^2 = 4 \), so \( r = \pm 2 \). It exists only where \( \cos 2\theta \ge 0 \), that is \( -\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{4} \) (and the half-turn opposite). \( r = \pm 2 \); \( |\theta| \le \dfrac{\pi}{4} \)

10. Classify \( r = 2 + 3\cos\theta \), \( r = 3 + 2\cos\theta \), \( r = 2 + 2\cos\theta \) and \( r = 4 + \cos\theta \).
    Show the full solution

    Compare \( a \) with \( b \): \( 2 \lt 3 \) inner loop; \( 2 \lt 3 \lt 4 \) dimpled; \( a = b \) cardioid; \( 4 \ge 2(1) \) convex. Inner loop; dimpled; cardioid; convex

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the modulus of \( 3 + 4i \).
   Show the full solution

   \( \sqrt{9 + 16} \). 5

2. Find the argument of \( 1 + i \).
   Show the full solution

   Equal parts, quadrant I. \( \dfrac{\pi}{4} \)

3. Write \( -2 \) in polar form.
   Show the full solution

   Modulus 2, on the negative real axis. \( 2\,\text{cis}\,\pi \)

4. Write \( 4\,\text{cis}\dfrac{\pi}{2} \) in rectangular form.
   Show the full solution

   \( 4(0 + i) \). \( 4i \)

5. Write \( i \) in polar form.
   Show the full solution

   \( \text{cis}\dfrac{\pi}{2} \)

6. Write \( -1 + i\sqrt3 \) in polar form.
   Show the full solution

   \( r = 2 \); quadrant II with reference angle \( \dfrac{\pi}{3} \). \( 2\,\text{cis}\dfrac{2\pi}{3} \)

7. Multiply \( 2\,\text{cis}\dfrac{\pi}{4} \) by \( 3\,\text{cis}\dfrac{\pi}{4} \).
   Show the full solution

   \( 6\,\text{cis}\dfrac{\pi}{2} \). \( 6i \)

8. Divide \( 8\,\text{cis}\,\pi \) by \( 2\,\text{cis}\dfrac{\pi}{3} \).
   Show the full solution

   \( 4\,\text{cis}\left(\pi - \dfrac{\pi}{3}\right) \). \( 4\,\text{cis}\dfrac{2\pi}{3} \)

9. Write \( 3 - 3i \) in polar form.
   Show the full solution

   \( r = 3\sqrt2 \); quadrant IV, argument \( -\dfrac{\pi}{4} \) (equivalently \( \dfrac{7\pi}{4} \)). Check: \( 3\sqrt2\cos\left(-\dfrac{\pi}{4}\right) = 3 \). \( 3\sqrt2\,\text{cis}\left(-\dfrac{\pi}{4}\right) \)

10. Show that multiplying by \( i \) rotates \( 1 + i \) by \( 90^\circ \), and say why multiplying moduli and adding arguments is reasonable.
    Show the full solution

    \( (1 + i)i = -1 + i \); the argument goes from \( \dfrac{\pi}{4} \) to \( \dfrac{3\pi}{4} \) and the modulus stays \( \sqrt2 \). The rule follows from the sum formulas: the real part of the product is \( r_1r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) = r_1r_2\cos(\theta_1 + \theta_2) \), and the imaginary part is \( r_1r_2(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2) = r_1r_2\sin(\theta_1 + \theta_2) \). The sum formulas give the rule

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Compute \( \left(\text{cis}\dfrac{\pi}{6}\right)^6 \).
   Show the full solution

   \( \text{cis}\,\pi \). \( -1 \)

2. Compute \( \left(2\,\text{cis}\dfrac{\pi}{3}\right)^3 \).
   Show the full solution

   \( 8\,\text{cis}\,\pi \). \( -8 \)

3. Compute \( (1 + i)^2 \).
   Show the full solution

   \( 1 + 2i + i^2 \). \( 2i \)

4. Compute \( (1 + i)^4 \).
   Show the full solution

   \( (2i)^2 \). \( -4 \)

5. Find the square roots of 1.
   Show the full solution

   \( \text{cis}\,0 \) and \( \text{cis}\,\pi \). \( 1 \) and \( -1 \)

6. Find the cube roots of 1.
   Show the full solution

   Angles \( 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \). \( 1 \), \( -\dfrac12 + \dfrac{\sqrt3}{2}i \), \( -\dfrac12 - \dfrac{\sqrt3}{2}i \)

7. Compute \( (\sqrt3 - i)^5 \).
   Show the full solution

   \( \sqrt3 - i = 2\,\text{cis}\left(-\dfrac{\pi}{6}\right) \). Then \( 32\,\text{cis}\left(-\dfrac{5\pi}{6}\right) = 32\left(-\dfrac{\sqrt3}{2} - \dfrac{i}{2}\right) \). \( -16\sqrt3 - 16i \)

8. Find the fourth roots of 16.
   Show the full solution

   Modulus \( 16^{1/4} = 2 \), angles \( 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} \). \( 2, 2i, -2, -2i \)

9. How far apart in angle are the sixth roots of unity, and what shape do they form?
   Show the full solution

   \( \dfrac{2\pi}{6} = \dfrac{\pi}{3} \), or \( 60^\circ \). Six equally spaced points on the unit circle: the vertices of a regular hexagon. \( 60^\circ \) apart; a regular hexagon

10. Show that the three cube roots of 1 add to zero, and explain why the \( n \)th roots of unity always do.
    Show the full solution

    \( 1 + \left(-\dfrac12 + \dfrac{\sqrt3}{2}i\right) + \left(-\dfrac12 - \dfrac{\sqrt3}{2}i\right) = 0 \). In general the \( n \)th roots of unity are the roots of \( x^n - 1 \), and the sum of the roots of a polynomial is minus the coefficient of \( x^{n-1} \) divided by the leading coefficient. The polynomial \( x^n - 1 \) has no \( x^{n-1} \) term when \( n \ge 2 \), so the sum is 0. Geometrically the roots are spaced evenly around the circle, so they balance about the origin like equal weights on a wheel. The coefficient of \( x^{n-1} \) is 0, and the roots balance

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( x = 2t + 1 \), \( y = t^2 \), find the point at \( t = 2 \).
   Show the full solution

   \( (5, 4) \). \( (5, 4) \)

2. For that curve, find \( t \) when \( x = 7 \).
   Show the full solution

   \( 2t + 1 = 7 \). \( t = 3 \)

3. Find the point at \( t = \dfrac{\pi}{2} \) on \( x = \cos t \), \( y = \sin t \).
   Show the full solution

   \( (0, 1) \)

4. Write parametric equations for the line through \( (0, 0) \) and \( (2, 4) \).
   Show the full solution

   Direction \( \langle 2, 4 \rangle \) from the origin. \( x = 2t \), \( y = 4t \)

5. Identify \( x = 5\cos t \), \( y = 5\sin t \).
   Show the full solution

   Distance 5 from the origin for every \( t \). A circle of radius 5

6. Tabulate \( x = t^2 - 1 \), \( y = t \) for \( t = -2, -1, 0, 1, 2 \).
   Show the full solution

   \( (3, -2), (0, -1), (-1, 0), (0, 1), (3, 2) \). These lie on a parabola opening to the right, with vertex \( (-1, 0) \). \( (3,-2), (0,-1), (-1,0), (0,1), (3,2) \)

7. Find the point at \( t = \dfrac{\pi}{6} \) on \( x = 2\cos t \), \( y = 3\sin t \).
   Show the full solution

   \( (2\cos 30^\circ,\ 3\sin 30^\circ) = (\sqrt3,\ 1.5) \). Check on the ellipse: \( \dfrac{3}{4} + \dfrac{2.25}{9} = 0.75 + 0.25 = 1 \) ✓. \( (1.732, 1.5) \)

8. Parametrize the segment from \( (1, 2) \) to \( (4, 8) \) and find its midpoint.
   Show the full solution

   \( x = 1 + 3t \), \( y = 2 + 6t \), \( 0 \le t \le 1 \). At \( t = 0.5 \): \( (2.5, 5) \). \( (2.5, 5) \)

9. How do \( x = \cos t \), \( y = \sin t \) and \( x = \cos 2t \), \( y = \sin 2t \) differ on \( 0 \le t \le 2\pi \)?
   Show the full solution

   Both lie on the unit circle. The second completes two laps in the same time, moving twice as fast, since its point at time \( t \) is at angle \( 2t \). Same path; the second is twice as fast

10. Explain why the parameter limits matter, using \( x = \cos t \), \( y = \sin t \) for \( 0 \le t \le \pi \).
    Show the full solution

    Over \( 0 \le t \le \pi \), \( y = \sin t \ge 0 \), so only the upper semicircle is traced, from \( (1, 0) \) to \( (-1, 0) \). The full circle needs \( 0 \le t \le 2\pi \). The equations alone do not say how much of the curve is drawn; the interval does. Only the upper half circle

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Eliminate \( t \): \( x = t + 1 \), \( y = 2t \).
   Show the full solution

   \( t = x - 1 \), so \( y = 2(x - 1) \). \( y = 2x - 2 \)

2. Eliminate \( t \): \( x = t^2 \), \( y = 2t \).
   Show the full solution

   \( t = \dfrac{y}{2} \), so \( x = \dfrac{y^2}{4} \). \( x = \dfrac{y^2}{4} \), a parabola opening right

3. Eliminate \( t \): \( x = 3\cos t \), \( y = 3\sin t \).
   Show the full solution

   \( x^2 + y^2 = 9(\cos^2 t + \sin^2 t) \). \( x^2 + y^2 = 9 \)

4. Eliminate \( t \): \( x = 2t \), \( y = 4t^2 \).
   Show the full solution

   \( t = \dfrac{x}{2} \), so \( y = 4 \cdot \dfrac{x^2}{4} \). \( y = x^2 \)

5. A ball is thrown upward at 14.7 m/s. When does it reach its peak?
   Show the full solution

   \( t = \dfrac{14.7}{9.8} \). 1.5 s

6. Eliminate \( t \): \( x = 4\cos t \), \( y = 2\sin t \).
   Show the full solution

   \( \dfrac{x^2}{16} + \dfrac{y^2}{4} = 1 \). An ellipse with semi-axes 4 and 2

7. A ball is thrown at 20 m/s at \( 40^\circ \) from the ground. Find the flight time and range.
   Show the full solution

   Vertical speed \( 20\sin 40^\circ = 12.856 \), horizontal \( 20\cos 40^\circ = 15.321 \). \( t = \dfrac{2(12.856)}{9.8} = 2.624 \) s; range \( 15.321 \times 2.624 = 40.2 \) m. Check with \( \dfrac{v_0^2\sin 80^\circ}{9.8} = \dfrac{400(0.98481)}{9.8} = 40.2 \). 2.62 s; 40.2 m

8. Find the maximum height of that ball.
   Show the full solution

   \( \dfrac{(12.856)^2}{2(9.8)} = \dfrac{165.28}{19.6} = 8.43 \). 8.43 m

9. Find the range of a ball thrown at 20 m/s at \( 50^\circ \), and compare with \( 40^\circ \).
   Show the full solution

   \( \dfrac{400\sin 100^\circ}{9.8} = \dfrac{400(0.98481)}{9.8} = 40.2 \) m, the same as at \( 40^\circ \), because \( \sin 100^\circ = \sin 80^\circ \). Complementary angles give equal ranges. At \( 45^\circ \) the range is \( \dfrac{400}{9.8} = 40.8 \) m, the maximum. 40.2 m, equal to the \( 40^\circ \) range

10. Two balls are thrown at 20 m/s at \( 30^\circ \) and \( 60^\circ \). Show they have the same range but different flight times and heights.
    Show the full solution

    Range: \( \dfrac{400\sin 60^\circ}{9.8} \) and \( \dfrac{400\sin 120^\circ}{9.8} \), both \( 35.3 \) m. Times: \( \dfrac{2(20)(0.5)}{9.8} = 2.04 \) s and \( \dfrac{2(20)(0.866)}{9.8} = 3.53 \) s. Heights: \( \dfrac{100}{19.6} = 5.10 \) m and \( \dfrac{300}{19.6} = 15.3 \) m. The higher throw stays up longer but moves forward more slowly. Same range 35.3 m; the \( 60^\circ \) throw takes longer and goes higher

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the vertex of \( x^2 = 4y \).
   Show the full solution

   \( (0, 0) \)

2. Find the focus of \( x^2 = 4y \).
   Show the full solution

   \( 4p = 4 \), so \( p = 1 \). \( (0, 1) \)

3. Find the focus and directrix of \( y^2 = 8x \).
   Show the full solution

   \( 4p = 8 \), \( p = 2 \); opens right. Focus \( (2, 0) \); directrix \( x = -2 \)

4. Which way does \( (x - 1)^2 = -8(y + 2) \) open, and where is its vertex?
   Show the full solution

   \( 4p = -8 \) is negative, so down. Vertex \( (1, -2) \). Down; \( (1, -2) \)

5. Write the parabola with vertex at the origin and focus \( (0, 3) \).
   Show the full solution

   \( p = 3 \), so \( x^2 = 12y \). \( x^2 = 12y \)

6. Write \( y = x^2 - 4x + 7 \) in the form \( (x - h)^2 = 4p(y - k) \) and find the focus.
   Show the full solution

   \( y - 7 = x^2 - 4x \), so \( y - 7 + 4 = (x - 2)^2 \), giving \( (x - 2)^2 = y - 3 \). \( 4p = 1 \), \( p = \dfrac14 \). Vertex \( (2, 3) \), focus \( \left(2, 3.25\right) \). Check at \( x = 2 \): \( y = 4 - 8 + 7 = 3 \) ✓. Vertex \( (2, 3) \); focus \( (2, 3.25) \)

7. Find the parabola with focus \( (3, 0) \) and directrix \( x = -3 \).
   Show the full solution

   Vertex at the origin, \( p = 3 \), opens right. \( y^2 = 12x \)

8. Find the width of \( x^2 = 8y \) at the focus.
   Show the full solution

   The focus is \( (0, 2) \). At \( y = 2 \): \( x^2 = 16 \), so \( x = \pm 4 \). Width 8, which equals \( |4p| = 8 \). 8

9. Verify that \( (4, 2) \) on \( x^2 = 8y \) is equidistant from the focus and the directrix.
   Show the full solution

   Focus \( (0, 2) \): distance \( \sqrt{16 + 0} = 4 \). Directrix \( y = -2 \): distance \( 2 - (-2) = 4 \). Both 4

10. A dish is modeled by \( y = \dfrac{x^2}{4} \) meters. Where should the receiver go, and why does a parabola work?
    Show the full solution

    \( x^2 = 4y \) has \( p = 1 \), so the focus is \( (0, 1) \), one meter above the vertex. A ray arriving parallel to the axis reflects through the focus; the parabola is the only curve that focuses every such ray at a single point. A circular dish would blur the signal (spherical aberration). At \( (0, 1) \); the parabola focuses parallel rays exactly

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( a \) and \( b \) for \( \dfrac{x^2}{16} + \dfrac{y^2}{9} = 1 \).
   Show the full solution

   Square roots of the denominators. \( a = 4 \), \( b = 3 \)

2. Find \( c \) for that ellipse.
   Show the full solution

   \( c^2 = 16 - 9 = 7 \). \( c = \sqrt7 \approx 2.646 \)

3. Find the foci of \( \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1 \).
   Show the full solution

   \( c^2 = 25 - 16 = 9 \), major axis horizontal. \( (\pm 3, 0) \)

4. Find the center of \( \dfrac{(x - 2)^2}{9} + \dfrac{(y + 1)^2}{4} = 1 \).
   Show the full solution

   Signs are opposite to those in the equation. \( (2, -1) \)

5. Find the length of the major axis of \( \dfrac{x^2}{49} + \dfrac{y^2}{25} = 1 \).
   Show the full solution

   \( 2a = 2(7) \). 14

6. Write the ellipse with vertices \( (\pm 5, 0) \) and co-vertices \( (0, \pm 3) \).
   Show the full solution

   \( a = 5 \), \( b = 3 \). \( \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 \)

7. Write the ellipse with foci \( (0, \pm 4) \) and major axis 10.
   Show the full solution

   Vertical major axis: \( a = 5 \), \( c = 4 \), \( b^2 = 25 - 16 = 9 \). \( \dfrac{y^2}{25} + \dfrac{x^2}{9} = 1 \)

8. Find the eccentricity of \( \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 \) and say what it measures.
   Show the full solution

   \( e = \dfrac{c}{a} = \dfrac45 = 0.8 \). It measures how far the ellipse is from a circle: 0 is a circle, and values near 1 are long and thin. 0.8

9. Find the area of \( \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 \).
   Show the full solution

   \( \pi ab = \pi(5)(3) = 15\pi \). A circle of radius \( r \) gives \( \pi r^2 \), the case \( a = b = r \). \( 15\pi \approx 47.12 \)

10. Explain what happens when \( a = b \), and describe how a gardener draws an ellipse with string and two pins.
    Show the full solution

    If \( a = b \) then \( c^2 = a^2 - b^2 = 0 \): the two foci coincide at the center and the curve is a circle, eccentricity 0. The gardener loops a string of length \( 2a + 2c \) around two pins at the foci and pulls it taut with a pencil. The two string segments always add to \( 2a \), which is the defining property, so the pencil traces the ellipse. A circle; the taut string enforces constant distance sum

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( a \) and \( b \) for \( \dfrac{x^2}{4} - \dfrac{y^2}{9} = 1 \).
   Show the full solution

   \( a = 2 \), \( b = 3 \)

2. Find \( c \) for that hyperbola.
   Show the full solution

   \( c^2 = 4 + 9 = 13 \). \( c = \sqrt{13} \approx 3.606 \)

3. Find the vertices.
   Show the full solution

   \( a \) from the center along the \( x \)-axis. \( (\pm 2, 0) \)

4. Find the asymptotes.
   Show the full solution

   \( y = \pm\dfrac{b}{a}x \). \( y = \pm\dfrac32 x \)

5. Find the foci of \( \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1 \).
   Show the full solution

   \( c^2 = 25 \). \( (\pm 5, 0) \)

6. Which way does \( \dfrac{y^2}{9} - \dfrac{x^2}{4} = 1 \) open, and what are its vertices?
   Show the full solution

   The positive term is \( y \): up and down. \( a = 3 \). Up and down; \( (0, \pm 3) \)

7. Write the hyperbola with vertices \( (\pm 3, 0) \) and foci \( (\pm 5, 0) \).
   Show the full solution

   \( a = 3 \), \( c = 5 \), \( b^2 = 25 - 9 = 16 \). \( \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1 \)

8. Find the asymptotes of \( \dfrac{y^2}{4} - \dfrac{x^2}{5} = 1 \).
   Show the full solution

   Up-down: slope \( \pm\dfrac{a}{b} = \pm\dfrac{2}{\sqrt5} \approx \pm 0.894 \). \( y = \pm\dfrac{2}{\sqrt5}x \)

9. Find the center and asymptote slopes of \( \dfrac{(x - 2)^2}{4} - \dfrac{(y + 1)^2}{9} = 1 \).
   Show the full solution

   Center \( (2, -1) \); slopes \( \pm\dfrac{b}{a} = \pm\dfrac32 \). \( (2, -1) \); \( \pm 1.5 \)

10. Compare ellipses and hyperbolas: the sign, the relation among \( a, b, c \), and the eccentricity.
    Show the full solution

    An ellipse has a plus between the squared terms and \( c^2 = a^2 - b^2 \), so \( c \lt a \) and \( 0 \le e \lt 1 \). A hyperbola has a minus and \( c^2 = a^2 + b^2 \), so \( c \gt a \) and \( e \gt 1 \). The ellipse closes up because the sum of distances is fixed; the hyperbola runs off because a fixed difference of distances can be kept indefinitely far away, approaching its asymptotes. Plus and \( a^2 - b^2 \) for ellipses; minus and \( a^2 + b^2 \) for hyperbolas

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Classify \( x^2 + y^2 = 9 \).
   Show the full solution

   Equal positive coefficients. Circle

2. Classify \( 4x^2 + y^2 = 16 \).
   Show the full solution

   Same sign, unequal. Ellipse

3. Classify \( x^2 - y^2 = 1 \).
   Show the full solution

   Opposite signs. Hyperbola

4. Classify \( y = x^2 + 1 \).
   Show the full solution

   Only \( x \) is squared. Parabola

5. Put \( x^2 + y^2 - 4x + 6y = 3 \) in standard form.
   Show the full solution

   Add 4 and 9: \( (x - 2)^2 + (y + 3)^2 = 16 \). Circle, center \( (2, -3) \), radius 4

6. Put \( 9x^2 + 4y^2 - 18x + 16y - 11 = 0 \) in standard form.
   Show the full solution

   \( 9(x^2 - 2x) + 4(y^2 + 4y) = 11 \), so \( 9(x - 1)^2 + 4(y + 2)^2 = 11 + 9 + 16 = 36 \). \( \dfrac{(x - 1)^2}{4} + \dfrac{(y + 2)^2}{9} = 1 \). Ellipse, center \( (1, -2) \), vertical major axis

7. Put \( x^2 - y^2 - 4x - 2y - 1 = 0 \) in standard form.
   Show the full solution

   \( (x^2 - 4x) - (y^2 + 2y) = 1 \), so \( (x - 2)^2 - 4 - (y + 1)^2 + 1 = 1 \), giving \( (x - 2)^2 - (y + 1)^2 = 4 \), or \( \dfrac{(x - 2)^2}{4} - \dfrac{(y + 1)^2}{4} = 1 \). Hyperbola, center \( (2, -1) \)

8. Put \( y^2 - 8x - 4y + 12 = 0 \) in standard form.
   Show the full solution

   \( y^2 - 4y = 8x - 12 \), so \( (y - 2)^2 = 8x - 12 + 4 = 8x - 8 = 8(x - 1) \). Parabola, vertex \( (1, 2) \), \( p = 2 \), opens right

9. Classify \( 4x^2 + 4y^2 = 36 \) and find its radius.
   Show the full solution

   Divide by 4: \( x^2 + y^2 = 9 \). Equal coefficients, so a circle of radius 3. Circle of radius 3

10. What do \( x^2 - y^2 = 0 \) and \( x^2 + y^2 = 0 \) describe, and why are they not the usual conics?
    Show the full solution

    \( x^2 - y^2 = 0 \) factors to \( (x - y)(x + y) = 0 \): the two lines \( y = \pm x \). And \( x^2 + y^2 = 0 \) has only the solution \( (0, 0) \): a single point. They are degenerate conics: the plane cuts the cone through its apex, instead of crossing it above or below. Such cases appear when the right side becomes zero after completing the square. Two lines; a single point

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the tenth term of \( 2, 5, 8, \dots \).
   Show the full solution

   \( 2 + 9(3) \). 29

2. Find the sum of the first 10 terms of \( 2, 5, 8, \dots \).
   Show the full solution

   \( \dfrac{10(2 + 29)}{2} \). 155

3. Find the sixth term of \( 2, 6, 18, \dots \).
   Show the full solution

   \( 2(3^5) = 2(243) \). 486

4. Find the sum of the first 5 terms of \( 2, 6, 18, \dots \).
   Show the full solution

   \( \dfrac{2(3^5 - 1)}{3 - 1} = 242 \). Check: \( 2 + 6 + 18 + 54 + 162 = 242 \). 242

5. Find \( 1 + 2 + 3 + \dots + 50 \).
   Show the full solution

   \( \dfrac{50 \cdot 51}{2} \). 1275

6. An arithmetic sequence has \( a_4 = 11 \) and \( a_9 = 31 \). Find \( a_1 \) and \( d \).
   Show the full solution

   \( 5d = 20 \), so \( d = 4 \), and \( a_1 = 11 - 12 = -1 \). Check: \( a_9 = -1 + 32 = 31 \). \( a_1 = -1 \), \( d = 4 \)

7. A geometric sequence has \( a_3 = 12 \) and \( a_6 = 96 \). Find \( a_1 \) and \( r \).
   Show the full solution

   \( r^3 = 8 \), \( r = 2 \), and \( a_1 = \dfrac{12}{4} = 3 \). \( a_1 = 3 \), \( r = 2 \)

8. Find the sum of the first 8 terms of \( 5, 10, 20, \dots \).
   Show the full solution

   \( 5(2^8 - 1) = 5(255) \). 1275

9. A saver deposits \$100 in month 1 and increases each deposit by \$10 each month. Find the total after 24 months.
   Show the full solution

   Arithmetic: \( a_{24} = 100 + 23(10) = 330 \). \( S_{24} = \dfrac{24(100 + 330)}{2} = 12(430) = 5160 \). \$5,160

10. Derive the arithmetic sum formula, and verify it for \( 1 + 2 + \dots + 10 \).
    Show the full solution

    Write \( S = a_1 + \dots + a_n \) and again in reverse. Adding the two lines term by term, each column is \( a_1 + a_n \), and there are \( n \) columns: \( 2S = n(a_1 + a_n) \). For \( 1 \) to \( 10 \): \( \dfrac{10(1 + 10)}{2} = 55 \), and adding directly: \( 1 + 2 + \dots + 10 = 55 \). \( S_n = \dfrac{n(a_1 + a_n)}{2} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( 1 + \tfrac13 + \tfrac19 + \cdots \).
   Show the full solution

   \( \dfrac{1}{1 - \frac13} = \dfrac{3}{2} \). 1.5

2. Find \( 4 + 2 + 1 + \cdots \).
   Show the full solution

   \( \dfrac{4}{1 - \frac12} \). 8

3. Does \( 3 + 6 + 12 + \cdots \) converge?
   Show the full solution

   \( r = 2 \), and \( |r| \ge 1 \). No

4. Write \( 0.\overline{5} \) as a fraction.
   Show the full solution

   \( \dfrac{0.5}{0.9} = \dfrac59 \). \( \dfrac59 \)

5. Find \( 10 - 5 + 2.5 - \cdots \).
   Show the full solution

   \( r = -\tfrac12 \): \( \dfrac{10}{1.5} = \dfrac{20}{3} \). \( 6.\overline{6} \)

6. Write \( 0.\overline{81} \) as a fraction.
   Show the full solution

   \( \dfrac{0.81}{0.99} = \dfrac{81}{99} = \dfrac{9}{11} \). \( \dfrac{9}{11} \)

7. For what \( x \) does \( 1 + x + x^2 + \cdots \) converge, and what is the sum at \( x = 0.2 \)?
   Show the full solution

   \( |x| \lt 1 \). The sum is \( \dfrac{1}{1 - x} \); at \( x = 0.2 \): \( \dfrac{1}{0.8} = 1.25 \). Check: \( 1 + 0.2 + 0.04 + 0.008 + \cdots = 1.2480\ldots \) ✓. \( |x| \lt 1 \); 1.25

8. A ball is dropped from 20 m and rebounds to 50 percent each time. Find the total distance.
   Show the full solution

   Rebounds: \( 10, 5, 2.5, \dots \): \( \dfrac{10}{0.5} = 20 \). Total \( 20 + 2(20) = 60 \). 60 m

9. Write \( 1.2\overline{4} = 1.2444\ldots \) as a fraction.
   Show the full solution

   \( 1.2 + 0.0444\ldots \). The repeating part: \( 0.04 + 0.004 + \cdots = \dfrac{0.04}{0.9} = \dfrac{4}{90} = \dfrac{2}{45} \). Total \( \dfrac65 + \dfrac{2}{45} = \dfrac{54 + 2}{45} = \dfrac{56}{45} \). Check: \( 56 \div 45 = 1.2444\ldots \) ✓. \( \dfrac{56}{45} \)

10. Explain how infinitely many terms can have a finite sum, using the partial sums of \( 1 + \tfrac12 + \tfrac14 + \cdots \).
    Show the full solution

    The partial sum after \( n \) terms is \( S_n = 2 - \dfrac{1}{2^{n-1}} \): 1, 1.5, 1.75, 1.875, ... Each new term closes half of the remaining gap to 2, so the sums rise toward 2 and never pass it. The infinite sum is defined as the number the partial sums approach, here 2. Terms that shrink geometrically add up to a finite amount; terms that shrink only like \( \dfrac1n \) do not. The partial sums approach 2

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( \displaystyle\sum_{k=1}^{4} k^2 \).
   Show the full solution

   \( 1 + 4 + 9 + 16 \). 30

2. Evaluate \( \displaystyle\sum_{k=1}^{100} k \).
   Show the full solution

   \( \dfrac{100 \cdot 101}{2} \). 5050

3. Evaluate \( \displaystyle\sum_{k=1}^{5} 3 \).
   Show the full solution

   Five terms of 3. 15

4. Evaluate \( \displaystyle\sum_{k=1}^{6}(2k - 1) \).
   Show the full solution

   \( 1 + 3 + 5 + 7 + 9 + 11 = 36 = 6^2 \). 36

5. Write \( 2 + 4 + 6 + 8 + 10 \) in sigma notation.
   Show the full solution

   \( \displaystyle\sum_{k=1}^{5} 2k \)

6. Evaluate \( \displaystyle\sum_{k=1}^{10} k^3 \).
   Show the full solution

   \( \left[\dfrac{10 \cdot 11}{2}\right]^2 = 55^2 \). 3025

7. Evaluate \( \displaystyle\sum_{k=1}^{20}(3k - 2) \).
   Show the full solution

   \( 3\sum k - 2(20) = 3(210) - 40 = 590 \). Check with the arithmetic formula: \( a_1 = 1 \), \( a_{20} = 58 \), \( \dfrac{20(59)}{2} = 590 \). 590

8. State the base case of the induction proof that \( 1 + 3 + \cdots + (2n - 1) = n^2 \) and verify it.
   Show the full solution

   \( n = 1 \): the left side is the single term 1 and the right side is \( 1^2 = 1 \). Both sides equal 1

9. Carry out the inductive step for that statement.
   Show the full solution

   Assume it for \( k \). The next term is \( 2k + 1 \), so the sum to \( k + 1 \) terms is \( k^2 + 2k + 1 = (k + 1)^2 \). The statement holds for \( k + 1 \)

10. Show that \( n^2 + n + 41 \) produces a prime for many \( n \) but not all, and explain what this says about checking cases.
    Show the full solution

    It gives 43, 47, 53, 61, ... for \( n = 1, 2, 3, 4 \), all prime, and continues to do so up to \( n = 39 \). At \( n = 40 \): \( 1600 + 40 + 41 = 1681 = 41^2 \), which is not prime. No number of confirming cases proves a general statement, because the failure may lie beyond them. Induction avoids the problem by proving the step from any \( n \) to \( n + 1 \), which covers infinitely many cases at once. 40 gives \( 41^2 \); patterns are not proofs

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \lim_{x \to 3}(2x + 1) \).
   Show the full solution

   The function is continuous, so substitute: \( 2(3) + 1 \). 7

2. Find \( \lim_{x \to 2} x^2 \).
   Show the full solution

   4

3. Estimate \( \lim_{x \to 3}\dfrac{x^2 - 9}{x - 3} \) from values at 2.99 and 3.01.
   Show the full solution

   At 2.99: \( \dfrac{8.9401 - 9}{-0.01} = 5.99 \). At 3.01: \( \dfrac{9.0601 - 9}{0.01} = 6.01 \). 6

4. Does \( \lim_{x \to 1}\dfrac{x^2 - 1}{x - 1} \) exist, although the function is undefined at 1?
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   For \( x \ne 1 \) the function equals \( x + 1 \), which approaches 2. Yes, it equals 2

5. If \( g(x) = x \) for \( x \ne 1 \) and \( g(1) = 5 \), find \( \lim_{x \to 1} g(x) \) and \( g(1) \).
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   Near 1 the outputs approach 1; the value at 1 is 5. Limit 1; value 5

6. Estimate \( \lim_{x \to 0}\dfrac{\sin x}{x} \).
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   At \( 0.1 \): 0.998334. At \( 0.01 \): 0.999983. 1

7. Estimate \( \lim_{x \to 0}\dfrac{\sqrt{x + 4} - 2}{x} \).
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   At 0.1: 0.24846. At 0.01: 0.249844. At \( -0.01 \): 0.250156. 0.25

8. Estimate \( \lim_{x \to 0}\dfrac{1 - \cos x}{x} \).
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   At 0.1: \( \dfrac{1 - 0.995004}{0.1} = 0.04996 \). At 0.01: 0.005. The values shrink toward 0. 0

9. What happens to \( \dfrac{|x - 1|}{x - 1} \) as \( x \to 1 \)?
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   For \( x \gt 1 \) it is \( +1 \); for \( x \lt 1 \) it is \( -1 \). The two sides disagree. The limit does not exist

10. Explain why the limit of a function at \( a \) need not equal \( f(a) \), and why we define the limit that way.
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    The limit records what the outputs do as the inputs approach \( a \), and never requires \( x = a \) itself. So \( f(a) \) may be undefined or may differ, as in \( g \) above. The definition is chosen so that limits can describe exactly the situations substitution cannot: a \( \dfrac00 \) such as \( \dfrac{x^2 - 4}{x - 2} \), and, in lesson 11.6, the slope of a tangent, which is a \( \dfrac00 \) by nature. The approach matters, not the arrival

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \lim_{x \to 2}(3x - 1) \).
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   5

2. Find \( \lim_{x \to 4}\dfrac{x^2 - 16}{x - 4} \).
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   \( x + 4 \to 8 \). 8

3. Find \( \lim_{x \to 0}\dfrac{x^2 + 5x}{x} \).
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   \( x + 5 \to 5 \). 5

4. Find \( \lim_{x \to 1}\dfrac{x^2 - 1}{x^2 + x - 2} \).
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   \( \dfrac{(x - 1)(x + 1)}{(x - 1)(x + 2)} = \dfrac{x + 1}{x + 2} \to \dfrac23 \). \( \dfrac23 \)

5. Find \( \lim_{x \to 3}\dfrac{x^2 - 9}{x^2 - x - 6} \).
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   \( \dfrac{(x - 3)(x + 3)}{(x - 3)(x + 2)} = \dfrac{x + 3}{x + 2} \to \dfrac65 \). \( \dfrac65 \)

6. Find \( \lim_{x \to 9}\dfrac{\sqrt{x} - 3}{x - 9} \).
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   \( x - 9 = (\sqrt x - 3)(\sqrt x + 3) \), so the quotient is \( \dfrac{1}{\sqrt x + 3} \to \dfrac16 \). \( \dfrac16 \)

7. Find \( \lim_{x \to 0}\dfrac{\frac{1}{x + 2} - \frac12}{x} \).
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   The numerator is \( \dfrac{2 - (x + 2)}{2(x + 2)} = \dfrac{-x}{2(x + 2)} \); dividing by \( x \) gives \( \dfrac{-1}{2(x + 2)} \to -\dfrac14 \). \( -\dfrac14 \)

8. Find \( \lim_{h \to 0}\dfrac{(2 + h)^2 - 4}{h} \).
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   \( \dfrac{4 + 4h + h^2 - 4}{h} = 4 + h \to 4 \). 4

9. Find \( \lim_{x \to 25}\dfrac{\sqrt{x} - 5}{x - 25} \).
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   \( \dfrac{1}{\sqrt x + 5} \to \dfrac{1}{10} \). \( \dfrac{1}{10} \)

10. Show that \( \dfrac00 \) can have different limits, using \( \dfrac{x}{x} \), \( \dfrac{x^2}{x} \) and \( \dfrac{x}{x^3} \) as \( x \to 0 \).
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    \( \dfrac{x}{x} = 1 \) for \( x \ne 0 \), limit 1. \( \dfrac{x^2}{x} = x \to 0 \). \( \dfrac{x}{x^3} = \dfrac{1}{x^2} \to \infty \), no finite limit. All three give \( \dfrac00 \) by substitution, so the form alone cannot decide; the algebra does. 1, 0, and no limit

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( f(x) = x + 1 \) when \( x \lt 2 \) and \( x^2 \) when \( x \ge 2 \), find \( \lim_{x \to 2^-} f(x) \).
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   Use the piece for \( x \lt 2 \): \( x + 1 \to 3 \). 3

2. For the same \( f \), find \( \lim_{x \to 2^+} f(x) \).
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   Use \( x^2 \): \( 4 \). 4

3. Does \( \lim_{x \to 2} f(x) \) exist?
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   \( 3 \ne 4 \). No

4. Find \( \lim_{x \to 0^+}\dfrac1x \).
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   Small positive \( x \) gives huge positive values. \( +\infty \)

5. Find \( \lim_{x \to 0^-}\dfrac1x \).
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   Small negative \( x \) gives huge negative values. \( -\infty \)

6. Does \( \lim_{x \to 3}\dfrac{|x - 3|}{x - 3} \) exist?
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   From the right \( +1 \); from the left \( -1 \). No

7. Describe \( \lim_{x \to 0}\dfrac{1}{x^2} \).
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   Both sides grow without bound and are positive. We write \( \infty \) to describe the behavior, but no number is approached. The values grow without bound: \( \infty \)

8. Find \( \lim_{x \to 2^+}\dfrac{x - 2}{|x - 2|} \).
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   For \( x \gt 2 \), \( |x - 2| = x - 2 \), so the quotient is 1. 1

9. For \( f(x) = 2x + 1 \) when \( x \lt 1 \) and \( 5 - x \) when \( x \ge 1 \), find both one-sided limits at 1 and decide whether the limit exists.
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   Left: \( 2(1) + 1 = 3 \). Right: \( 5 - 1 = 4 \). They differ. 3 and 4; the limit does not exist

10. Name the three ways a limit can fail to exist and give an example of each.
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    A jump: \( \dfrac{|x|}{x} \) at 0, with one-sided limits \( \pm 1 \). Infinite behavior: \( \dfrac1x \) at 0, growing without bound. Oscillation: \( \sin\dfrac1x \) at 0, swinging between \( -1 \) and 1 infinitely often. In the first the one-sided limits exist and differ; in the second they are infinite; in the third they do not exist. Jump, blow-up, oscillation