Lesson 1.1 · Unit 1 · F-IF.1-2
A rule that gives each input exactly one output
Every topic in this course is a function, so the language has to be exact. A function is not a formula. It is a rule, and the formula is only one way of writing it down. Before any computation, ask what inputs the rule accepts and what outputs it can produce, because those two sets, the domain and the range, are where calculus later goes looking for trouble.
- A function assigns to each input exactly one output. Two different inputs may share an output; one input may never have two.
- \( f(a) \) means the output when the input is \( a \). It is not \( f \) times \( a \). Replace every \( x \) in the rule by \( a \), using parentheses.
- The domain is every input the rule can accept. Start from all real numbers and remove what breaks the rule.
- Two things break a rule at this level: a zero in a denominator, and a negative number under an even root.
- The range is every output the rule produces. For a quadratic, find the vertex; for other rules, think about what the expression can and cannot equal.
- Vertical line test: if any vertical line meets a graph twice, one input has two outputs and the graph is not a function.
- Write domain and range in interval notation, using \( \cup \) to join pieces and round brackets at values that are excluded.
- Substitute an expression exactly as written: \( f(a + 1) \) puts the whole \( a + 1 \) in every place where \( x \) stood.
Where students lose marks: the linearity error. It is tempting to write \( f(a + b) = f(a) + f(b) \), but for \( f(x) = x^2 \) we have \( f(1 + 2) = 9 \) while \( f(1) + f(2) = 5 \). A function does not distribute over a sum unless it is linear, and almost none of this course's functions are.
The problem. (a) For \( f(x) = 2x^2 - 3x + 1 \), find \( f(-2) \) and \( f(a + 1) \). (b) Find the domain of \( g(x) = \dfrac{\sqrt{x - 3}}{x - 5} \). (c) Find the range of \( h(x) = x^2 + 2x + 5 \). (d) Decide whether \( x^2 + y^2 = 25 \) defines \( y \) as a function of \( x \).
Step one: evaluate \( f(-2) \) for (a). Put \( -2 \) in every place \( x \) stood, inside parentheses so the sign is squared correctly: \( f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 15 \).
Step two: evaluate \( f(a + 1) \). The input is now an expression: \( f(a + 1) = 2(a + 1)^2 - 3(a + 1) + 1 \). Expand: \( 2(a^2 + 2a + 1) - 3a - 3 + 1 = 2a^2 + 4a + 2 - 3a - 2 = 2a^2 + a \). Check with \( a = 1 \): \( f(2) = 8 - 6 + 1 = 3 \), and \( 2(1) + 1 = 3 \). Correct.
Step three: list the restrictions for (b). There is an even root, so \( x - 3 \ge 0 \), which gives \( x \ge 3 \). There is a denominator, so \( x - 5 \ne 0 \), which gives \( x \ne 5 \).
Step four: combine them. Both restrictions apply at once. Take \( x \ge 3 \) and remove 5: domain \( [3, 5) \cup (5, \infty) \). The square bracket at 3 is included, because \( \sqrt{0} = 0 \) is fine. The round bracket at 5 is excluded, because the denominator would be zero.
Step five: find the range for (c). Complete the square: \( x^2 + 2x + 5 = (x + 1)^2 + 4 \). A square is never negative, so the smallest value is 4, reached at \( x = -1 \). There is no largest value. Range \( [4, \infty) \). Check: \( h(0) = 5 \ge 4 \) and \( h(-1) = 1 - 2 + 5 = 4 \). Correct.
Step six: solve (d) for \( y \). \( y^2 = 25 - x^2 \), so \( y = \pm\sqrt{25 - x^2} \).
Step seven: test a single input. Take \( x = 3 \). Then \( y = \pm 4 \), and both \( (3, 4) \) and \( (3, -4) \) satisfy the equation. One input has two outputs.
Step eight: state the conclusion and what rescues it. The circle fails the vertical line test, so \( y \) is not a function of \( x \). Restricting to the upper half, \( y = \sqrt{25 - x^2} \), does define a function, with domain \( [-5, 5] \) and range \( [0, 5] \). This move, cutting a graph down to a piece that passes the test, is how inverse functions and inverse trigonometric functions are made later in the course.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- If \( f(x) = 3x - 7 \), find \( f(4) \).
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\( 3(4) - 7 = 12 - 7 \). 5
- 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
- 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 \( (-\infty, -2) \cup (-2, \infty) \)
- Find the domain of \( f(x) = \sqrt{x + 5} \).
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Require \( x + 5 \ge 0 \). \( [-5, \infty) \)
- Is \( \{(1, 2), (3, 4), (1, 5)\} \) a function?
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The input 1 is paired with both 2 and 5. No
- 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. \( 2xh + h^2 \)
- 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 \( x \ne 5 \). \( [3, 5) \cup (5, \infty) \)
- 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 \( x = 0 \): \( f(0) = -4 \le 5 \). \( (-\infty, 5] \)
- 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. \( (-\infty, -3) \cup (3, \infty) \)
- 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 \( f(x) = mx \)