AP Precalculus
AP Precalc 2.7–2.9: Free Response
Shared by Ashton · Nov 12, 2026 · 0 plays
Two AP-style FRQs for Concept Check 2.7–2.9: a no-calculator one on compositions and a rational inverse, and a calculator-active one modeled on the class FRQ 2 example (composition, solving, end behavior, justifying invertibility). Every part has a rubric and a full-credit sample answer.
7 questions across 2 sections. Write your answer, then check it against the rubric and score yourself.
Sample questions
Open an item to see its answer.
(a) (i) Find f(g(1)). (ii) Find (g ∘ f)(3). Show the work that leads to each answer.
Sample answer: (i) g(1) = 0, so f(g(1)) = f(0) = (0 + 1)/(0 − 2) = −1/2. (ii) f(3) = (3 + 1)/(3 − 2) = 4, so (g ∘ f)(3) = g(4) = 5.
(b) Find (h ∘ f)(5) and (f ∘ h)(4), or indicate that a value is not defined. Give a reason for any value that is not defined.
Sample answer: (h ∘ f)(5) = h(f(5)) = h(2) = log₂2 = 1. (f ∘ h)(4) = f(h(4)) = f(2) = 3/0, which is not defined because x = 2 is not in the domain of f (the denominator is zero).
(c) (i) Find an expression for f⁻¹(x). (ii) State the domain of f⁻¹ and explain how it relates to f.
Sample answer: (i) x = (y + 1)/(y − 2) → x(y − 2) = y + 1 → xy − 2x = y + 1 → xy − y = 2x + 1 → y(x − 1) = 2x + 1 → f⁻¹(x) = (2x + 1)/(x − 1). (ii) The domain of f⁻¹ is all real numbers except x = 1. The domain of f⁻¹ is the range of f, and f never outputs 1 (horizontal asymptote y = 1).
(d) Verify that f⁻¹(f(x)) = x for x ≠ 2.
Sample answer: f⁻¹(f(x)) = [2(x + 1)/(x − 2) + 1] / [(x + 1)/(x − 2) − 1]. Numerator: (2x + 2 + x − 2)/(x − 2) = 3x/(x − 2). Denominator: (x + 1 − x + 2)/(x − 2) = 3/(x − 2). Dividing: (3x/(x − 2)) · ((x − 2)/3) = x.
Part A: (i) Find h(2), or indicate that it is not defined. (ii) Find h(5), or indicate that it is not defined. Show the work that leads to each answer.
Sample answer: (i) h(2) = g(f(2)) = g(0) = 6/(0 − 2) = −3. (ii) h(5) = g(f(5)) = g(2) = 6/(2 − 2), which is not defined because g has a vertical asymptote at x = 2.