AP Precalculus: Inverse Functions (Drill 1)

Question 1. The table below gives selected values of a one-to-one function h.

x	−2	0	3	5
h(x)	7	1	−4	0

What is the value of h⁻¹(1)?

• A) −2
• B) 0 ✓
• C) 1
• D) 5

Explanation: Choice B is correct. h⁻¹(1) asks: for what x-value does h(x) = 1? From the table, h(0) = 1, so h⁻¹(1) = 0. The inverse function swaps inputs and outputs: if h(0) = 1, then h⁻¹(1) = 0. Choice A is incorrect because h(−2) = 7, not 1; the student may have read the first column of the table without checking which output matches 1. Choice C is incorrect because 1 is the output value we are given as input to h⁻¹, not the answer; the student confused finding h⁻¹(1) with simply locating 1 in the table. Choice D is incorrect because h(5) = 0, not 1; the student confused h⁻¹(1) with h⁻¹(0), which equals 5.

Question 2. If f(x) = \(\dfrac{2x + 6}{3}\), which of the following defines f⁻¹(x)?

• A) \(\dfrac{3x - 6}{2}\) ✓
• B) \(\dfrac{3x + 6}{2}\)
• C) \(\dfrac{3}{2x + 6}\)
• D) \(\dfrac{2x - 6}{3}\)

Explanation: Choice A is correct. To find f⁻¹(x), write y = (2x + 6)/3, then swap x and y: x = (2y + 6)/3. Solve for y: 3x = 2y + 6, so 2y = 3x − 6, giving y = (3x − 6)/2. Therefore f⁻¹(x) = (3x − 6)/2. Choice B is incorrect because (3x + 6)/2 results from a sign error when solving 2y = 3x − 6; the student added 6 instead of subtracting it. Choice C is incorrect because 3/(2x + 6) is the reciprocal of f(x), not its inverse; the student confused "inverse function" with "multiplicative inverse (reciprocal)." Choice D is incorrect because (2x − 6)/3 has the same structure as f(x) with the sign of the constant changed; the student never swapped x and y and simply altered the original formula.

Question 3. Which of the following best justifies that f(x) = 3x − 5 and g(x) = \(\dfrac{x+5}{3}\) are inverse functions?

• A) f(0) = −5 and g(0) = 5/3, and these values are symmetric about the x-axis.
• B) The slope of f is 3 and the slope of g is 1/3, and 3 × (1/3) = 1.
• C) f(g(x)) = 3 · (x+5)/3 − 5 = x and g(f(x)) = (3x − 5 + 5)/3 = x, so both compositions equal x. ✓
• D) f(5) = 10 and g(10) = 5, showing the functions swap one input-output pair.

Explanation: Choice C is correct. Two functions are inverses if and only if f(g(x)) = x AND g(f(x)) = x for all x in their domains. Computing f(g(x)): f((x+5)/3) = 3·(x+5)/3 − 5 = (x + 5) − 5 = x. Computing g(f(x)): g(3x − 5) = (3x − 5 + 5)/3 = 3x/3 = x. Both compositions yield the identity, which is the definition-based justification required by Skill 3.C. Choice A is incorrect because comparing output values at a single point says nothing about the general inverse relationship; many non-inverse function pairs share specific output values. Choice B is incorrect because the product of slopes equaling 1 is a property of perpendicular linear functions, not inverse functions; non-linear inverses do not have this slope relationship at all. Choice D is incorrect because verifying that one input-output pair is swapped is insufficient; a single example does not prove a general inverse relationship, as many non-inverse functions could satisfy one such pair.

Question 4. A rideshare app calculates the fare F (in dollars) for a trip of d miles using F(d) = 1.5d + 2.50. A driver wants to determine the trip distance d from a given fare F. Which of the following correctly expresses d as a function of F?

• A) d = (F − 2.50) ÷ 1.5 ✓
• B) d = 1.5F + 2.50
• C) d = (F + 2.50) ÷ 1.5
• D) d = F ÷ 1.5

Explanation: Choice A is correct. Starting from F = 1.5d + 2.50, solve for d: subtract 2.50 from both sides to get F − 2.50 = 1.5d, then divide by 1.5 to get d = (F − 2.50)/1.5. As a check: a $14.50 fare gives d = (14.50 − 2.50)/1.5 = 12/1.5 = 8 miles, and verifying: F(8) = 1.5(8) + 2.50 = 14.50 ✓. Choice B is incorrect because d = 1.5F + 2.50 applies the same structure as the original formula with F as input; the student used F exactly as d appeared in the original, which gives the distance formula the same form rather than inverting it. Choice C is incorrect because (F + 2.50)/1.5 results from adding rather than subtracting the base fare; a sign error when isolating the term containing d. Choice D is incorrect because d = F/1.5 omits subtraction of the $2.50 base fare; the student divided by the per-mile rate without first removing the fixed charge.

Question 5. The function f(x) = x² − 4 does not have an inverse on its natural domain. If the domain of f is restricted to x ≥ 0, which of the following defines f⁻¹(x) and states its domain correctly?

• A) f⁻¹(x) = √x + 4, domain x ≥ −4 ✓
• B) f⁻¹(x) = √x + 4, domain x ≥ 0
• C) f⁻¹(x) = √x − 4, domain x ≥ 4
• D) f⁻¹(x) = √x − 4, domain x ≥ 0

Explanation: Choice A is correct. To find f⁻¹, write y = x² − 4 and swap variables: x = y² − 4 (with y ≥ 0 from the restricted domain). Solving for y: y² = x + 4, so y = √(x + 4) (positive root only, since y ≥ 0). The domain of f⁻¹ equals the range of f on [0, ∞). When x ≥ 0, f(x) = x² − 4 ≥ −4, so the range of f is [−4, ∞), which becomes the domain of f⁻¹. Choice B is incorrect because the domain x ≥ 0 is the restricted domain of the original f, not of f⁻¹; the domain of the inverse equals the range of the original function, which begins at −4, not 0. Choice C is incorrect because √(x − 4) comes from subtracting 4 instead of adding it when solving y² = x + 4; the student made a sign error during the algebraic isolation of y². Choice D is incorrect because √x − 4 results from incorrectly taking the square root of x first and then subtracting 4, rather than correctly solving y² = x + 4 before taking the root.