Lesson 1.1 · Unit 1 · F-IF.1, F-IF.2
What a function is, and what its notation actually says
Algebra 2 is organized around functions rather than around equations, so the year starts by pinning down what a function is and what the notation means. Most of the trouble students have with \( f(x) \) later comes from reading it as multiplication early.
- A relation pairs inputs with outputs. It is a function when every input has exactly one output.
- The vertical line test reads that condition off a graph: if any vertical line meets the graph twice, some input has two outputs and it is not a function.
- \( f(x) \) is not multiplication. It names the output of the function \( f \) at the input \( x \), and the parentheses are part of the name.
- To evaluate, replace every \( x \) with the input, in brackets, then simplify.
- The input can be an expression, so \( f(a + 1) \) means substitute \( a + 1 \) everywhere \( x \) appeared.
- The domain is the set of allowed inputs. Unless stated, it is every real number the rule can actually accept.
- Three things exclude an input: a zero denominator, a negative radicand under an even root, and a nonpositive argument of a logarithm.
- The range is the set of outputs produced, usually read from the graph rather than computed.
Where students lose marks: giving a domain as a list of excluded values when the question asked for the domain. "All reals except 7" is an answer; "7" is not. Write it as an interval or as a condition, and say what it is.
The problem. Let \( f(x) = 3x^2 - 5x + 2 \). (a) Find \( f(-2) \). (b) Find \( f(a + 1) \) and simplify. (c) Find the domain of \( g(x) = \dfrac{\sqrt{x - 3}}{x - 7} \). (d) Explain why the vertical line test works.
Step one: evaluate (a) with brackets. Substitute \( -2 \) for every \( x \), keeping it in brackets so the squaring applies to the sign: \( f(-2) = 3(-2)^2 - 5(-2) + 2 \).
Step two: simplify. \( 3(4) + 10 + 2 = 12 + 10 + 2 = 24 \). Note the middle term: \( -5(-2) = +10 \). Dropping that sign change is the most common slip in evaluation, and writing the brackets prevents it.
Step three: substitute for (b). \( f(a + 1) = 3(a + 1)^2 - 5(a + 1) + 2 \). Every \( x \) became \( (a + 1) \), including the one inside the square.
Step four: expand and simplify (b). \( (a + 1)^2 = a^2 + 2a + 1 \), so \( 3(a^2 + 2a + 1) = 3a^2 + 6a + 3 \). And \( -5(a + 1) = -5a - 5 \). Together: \( 3a^2 + 6a + 3 - 5a - 5 + 2 = 3a^2 + a \).
Step five: check (b). Put \( a = 1 \). The formula gives \( 3(1) + 1 = 4 \). Directly, \( f(1 + 1) = f(2) = 3(4) - 10 + 2 = 4 \). They agree. Checking a simplified expression at one value costs seconds and catches most algebra errors.
Step six: find the restrictions for (c). Two things can go wrong. The square root is an even root, so its radicand must be nonnegative: \( x - 3 \ge 0 \), giving \( x \ge 3 \). The denominator cannot be zero: \( x - 7 \ne 0 \), giving \( x \ne 7 \).
Step seven: combine and write (c) as an interval. Both conditions must hold at once, so the domain is every \( x \) at least 3 except 7: \( [3, 7) \cup (7, \infty) \). The bracket at 3 is square because 3 is included, and both sides of 7 are open because 7 is excluded.
Step eight: answer (d). A point on a graph is a pair of an input and its output, with the input read horizontally. All points on one vertical line share the same \( x \)-coordinate, that is the same input. So two points on one vertical line mean one input with two different outputs, which is exactly what the definition of a function forbids. The test is the definition restated geometrically, not a separate rule. A useful consequence: a circle is not a function, and neither is \( x = y^2 \), since both fail the test. Relations are not useless; they simply cannot be written as \( y = f(x) \) without splitting them into pieces.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use \( f(x) = x^2 - 4x \) where a function is not named.
- Find \( f(3) \).
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\( 9 - 12 \). \( -3 \)
- Find \( f(-1) \).
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\( 1 + 4 \). 5
- Find \( f(0) \).
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0
- Find the domain of \( \dfrac{1}{x - 5} \).
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Only the zero denominator excludes anything. All reals except 5
- Find the domain of \( \sqrt{x + 2} \).
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\( x + 2 \ge 0 \). \( x \ge -2 \)
- Find \( f(t - 2) \) and simplify.
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\( (t - 2)^2 - 4(t - 2) = t^2 - 4t + 4 - 4t + 8 = t^2 - 8t + 12 \). Check at \( t = 3 \): the formula gives \( 9 - 24 + 12 = -3 \), and \( f(1) = 1 - 4 = -3 \). Agrees. \( t^2 - 8t + 12 \)
- Find the domain of \( \dfrac{x + 1}{x^2 - 9} \).
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Factor the denominator: \( (x - 3)(x + 3) \), which is zero at \( x = 3 \) and \( x = -3 \). Domain: all reals except 3 and \( -3 \), or \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \). Note that the numerator plays no part: a zero numerator gives an output of zero, which is perfectly allowed. All reals except \( \pm 3 \)
- Find the domain of \( \dfrac{\sqrt{2x + 6}}{x - 1} \).
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Radicand nonnegative: \( 2x + 6 \ge 0 \), so \( x \ge -3 \). Denominator nonzero: \( x \ne 1 \). Both must hold: \( [-3, 1) \cup (1, \infty) \). Check a boundary: at \( x = -3 \) the expression is \( \dfrac{0}{-4} = 0 \), which is fine, so \( -3 \) is included. \( [-3, 1) \cup (1, \infty) \)
- Explain why \( f(x) \) is not \( f \) times \( x \), and why the distinction matters.
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\( f \) is not a number. It is a rule, and the notation \( f(x) \) names the output that rule produces from the input \( x \). The parentheses are part of the naming, not a multiplication sign. Why it matters. Reading it as multiplication makes several later statements come out wrong. A student who believes \( f(x) \) is a product will "simplify" \( \dfrac{f(2x)}{f(x)} \) to 2, which is almost never true. Take \( f(x) = x^2 \): then \( f(2x) = 4x^2 \) and the ratio is 4, not 2. Take \( f(x) = x + 1 \): the ratio is \( \dfrac{2x + 1}{x + 1} \), which is not constant at all. The same misreading produces \( f(a + b) = f(a) + f(b) \), which is the linearity error this course names. It holds only for functions of the form \( f(x) = mx \), and for nothing else. The reliable habit. Read \( f(x) \) aloud as "f of x" rather than "f x". It costs nothing and it keeps the two ideas apart. \( f \) is a rule, not a factor; reading it as a product produces the linearity error
- A function is defined by the table: \( f(1) = 4 \), \( f(2) = 7 \), \( f(3) = 4 \),
\( f(4) = 10 \). State its domain and range, and say whether \( f \) has an inverse
function.
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Domain. The set of inputs the table defines: \( \{1, 2, 3, 4\} \). Range. The set of outputs produced, each listed once: \( \{4, 7, 10\} \). The value 4 appears twice in the table but the range is a set, so it is written once. Is it a function? Yes. Each input appears exactly once with exactly one output. Two inputs sharing an output is entirely allowed; it is two outputs sharing an input that is forbidden. Does it have an inverse function? No. An inverse would have to send 4 back to its input, but 4 came from both 1 and 3. The inverse relation would pair the input 4 with two different outputs, which is not a function. The general statement. A function has an inverse function exactly when no output is repeated, which is the horizontal line test of lesson 1.7. This table fails it. What could be done. Restricting the domain to \( \{1, 2, 4\} \), or to \( \{2, 3, 4\} \), removes the repetition and produces a function that does have an inverse. That is the same move that makes \( \sqrt{x} \) the inverse of \( x^2 \). Domain \( \{1,2,3,4\} \), range \( \{4,7,10\} \), no inverse function because 4 is repeated