Algebra 2 Composition and Inverse Functions Test Solutions and Scoring (PDF)

Algebra 2 math test solutions: Composition and Inverse Functions – Maths PDF

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This composition inverses test answers for Algebra 2 covers composition and inverse functions (Common Core F-BF.A.1c, F-BF.B.4a), with a free PDF to print.

Use this answer key and scoring guide to grade the test on composition and inverse functions (F-BF.A.1c). Each solution shows the full reasoning, and the scoring note explains how many points to award for correct setup, correct work, and a correct final answer. Partial credit rewards a sound method even when a small arithmetic slip changes the result.

Algebra 2 · Common Core State Standards: F-BF.A.1c, F-BF.B.4a

Composition inverses test answers: worked solutions

Question 1 – Compose from tables [2 pts]

\(g(1) = 0\), so \(f(g(1)) = f(0) = 2\). \(f(2) = 1\), so \(g(f(2)) = g(1) = 0\).

f(g(1)) = 2; g(f(2)) = 0

Scoring: 1 point for each value.

Question 2 – Compose expressions [3 pts]

\(f(g(x)) = (2x + 3)^{2} – 1 = 4x^{2} + 12x + 9 – 1 = 4x^{2} + 12x + 8\).

\(g(f(x)) = 2(x^{2} – 1) + 3 = 2x^{2} + 1\).

4x² + 12x + 8 and 2x² + 1; not equal

Scoring: 1 point for each composition, 1 point for the comparison.

Question 3 – Domain [2 pts]

\(f(g(x)) = \sqrt{x – 5}\). The radicand must be nonnegative: \(x \ge 5\).

\(\sqrt{x – 5}\), domain \(x \ge 5\)

Scoring: 1 point for the composition, 1 point for the domain.

Question 4 – Linear inverse [2 pts]

Swap: \(x = \frac{y – 4}{5}\), so \(y = 5x + 4\).

\(f^{-1}(x) = 5x + 4\)

Scoring: 1 point for swapping, 1 point for solving.

Question 5 – Rational inverse [3 pts]

Swap: \(x = \frac{2y + 1}{y – 3}\). Multiply: \(xy – 3x = 2y + 1\).

Collect \(y\) terms: \(xy – 2y = 3x + 1\), so \(y(x – 2) = 3x + 1\).

\(f^{-1}(x) = \frac{3x + 1}{x – 2}\)

Scoring: 1 point for swapping and clearing the fraction, 1 point for collecting the y-terms, 1 point for the inverse.

Question 6 – Verify [2 pts]

\(f(g(x)) = \sqrt[3]{x^{3} – 7 + 7} = x\). \(g(f(x)) = (\sqrt[3]{x + 7})^{3} – 7 = x\).

Both compositions give x

Scoring: 1 point for each composition.

Question 7 – Graph of an inverse [2 pts]

(4, 3) and (3, 4) are mirror images.

Swapping coordinates gives \((3, 4)\).

\(x = \frac{y}{2} + 1\) gives \(y = 2x – 2\).

(3, 4); f^(-1)(x) = 2x – 2

Scoring: 1 point for the point, 1 point for the inverse.

Question 8 – Restricted domain [2 pts]

Inverse: √(x + 1).

\(x = y^{2} – 1\), so \(y = \sqrt{x + 1}\) (nonnegative because \(x \ge 0\) originally).

Domain: \(x \ge -1\).

\(f^{-1}(x) = \sqrt{x + 1}\), \(x \ge -1\)

Scoring: 1 point for the inverse, 1 point for the domain.

Question 9 – Find the error [2 pts]

\(f(g(x)) = 2x + 1\) but \(g(f(x)) = 2(x + 1) = 2x + 2\).

2x + 1 is not equal to 2x + 2

Scoring: 1 point for each composition.

Total score

Total

20 points

Key facts behind this composition inverses test answers

Key facts

f(g(x)): apply g first, then f.
Composition order matters.
Inverse: swap x and y, then solve for y.
f(f^(-1)(x)) = x and f^(-1)(f(x)) = x.
Graphs of f and f^(-1) are mirror images over y = x.

How to use this composition inverses test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

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