Algebra 2 Composition and Inverse Functions Worksheet Solutions (PDF)

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

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This composition inverses 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.

Every worksheet problem on composition and inverse functions is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where composing in the wrong order or forgetting to swap x and y can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

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

Composition inverses answers: exercises 1 to 10

Exercise 1 – Evaluate a composition

$g(2) = 4$, so $f(g(2)) = f(4) = 11$. $f(2) = 7$, so $g(f(2)) = g(7) = 49$.

f(g(2)) = 11 and g(f(2)) = 49

Exercise 2 – Compose expressions

$f(g(x)) = f(3x) = 3x + 4$.

$g(f(x)) = g(x + 4) = 3(x + 4) = 3x + 12$.

3x + 4 and 3x + 12; not equal

Exercise 3 – Compose from tables

f(g(x)) for x = 1, 2, 3.

$g(1) = 4$, $f(4) = 7$. $g(2) = 1$, $f(1) = 5$. $g(3) = 2$, $f(2) = 0$.

7, 5, 0

Exercise 4 – Composition with a radical

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

$\sqrt{x^{2}} = |x|$. The domain is all real numbers, because $x^{2} + 1 \ge 1$.

$f(g(x)) = |x|$

Exercise 5 – Domain of a composition

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

The denominator is 0 when $x = -1$.

$f(g(x)) = \frac{1}{x + 1}$, $x \ne -1$

Exercise 6 – Inverse from a table

The outputs rise by 3 each step: slope 3. $f(1) = 7$ gives $f(x) = 3x + 4$.

Swap: $x = 3y + 4$, so $y = \frac{x – 4}{3}$.

$f^{-1}(25) = \frac{25 – 4}{3} = 7$.

$f(x) = 3x + 4$; $f^{-1}(x) = \frac{x – 4}{3}$; input 7

Exercise 7 – Inverse with a fraction

Swap: $x = \frac{y + 5}{2}$. Solve: $y = 2x – 5$.

Check: $f(2x – 5) = \frac{2x – 5 + 5}{2} = x$.

$f^{-1}(x) = 2x – 5$

Exercise 8 – Inverse of a cube

Swap: $x = y^{3} + 1$, so $y^{3} = x – 1$.

$f^{-1}(x) = \sqrt[3]{x – 1}$

Exercise 9 – Inverse of a rational function

Swap: $x = \frac{2}{y – 1} + 3$. Subtract 3: $x – 3 = \frac{2}{y – 1}$.

So $y – 1 = \frac{2}{x – 3}$ and $y = \frac{2}{x – 3} + 1$.

$f^{-1}(x) = \frac{2}{x – 3} + 1$

Exercise 10 – Verify inverses

$f(g(x)) = 4 \cdot \frac{x + 8}{4} – 8 = x + 8 – 8 = x$.

$g(f(x)) = \frac{4x – 8 + 8}{4} = x$.

Both compositions equal x, so they are inverses

Exercises 11 and up

Exercise 11 – Read the graph

(2, 3) on f gives (3, 2) on f^(-1).

$f(2) = 3$, so $(2, 3)$ is on $f$ and $(3, 2)$ is on $f^{-1}$.

Algebra: $f^{-1}(x) = \frac{x + 1}{2}$, so $f^{-1}(3) = 2$.

$f^{-1}(3) = 2$

Exercise 12 – Inverse of a table

Swap the rows.

Swap inputs and outputs. The row of outputs of f becomes the inputs of the inverse.

$f^{-1}(2) = 1$

Exercise 13 – Restricted quadratic

Swap: $x = y^{2} + 2$, so $y^{2} = x – 2$. Since the original domain is $x \ge 0$, the inverse’s outputs must be nonnegative: $y = \sqrt{x – 2}$.

The domain of the inverse equals the range of $f$: $x \ge 2$.

$f^{-1}(x) = \sqrt{x – 2}$, domain $x \ge 2$

Exercise 14 – Why no inverse?

$f(2) = 4$ and $f(-2) = 4$. A horizontal line at $y = 4$ meets the graph twice, so $f$ is not one-to-one.

An inverse would have to send 4 to both 2 and -2, which a function cannot do.

Not one-to-one; restrict the domain to x ≥ 0 (or x ≤ 0)

Exercise 15 – Vertex form with a restriction

Swap: $x = (y – 3)^{2} – 1$, so $(y – 3)^{2} = x + 1$.

Because $y \ge 3$, take the positive root: $y – 3 = \sqrt{x + 1}$.

$f^{-1}(x) = \sqrt{x + 1} + 3$, domain $x \ge -1$

Exercise 16 – Temperature

Solve for $C$: $F – 32 = \frac{9}{5}C$, so $C = \frac{5}{9}(F – 32)$.

$C = \frac{5}{9}(86 – 32) = \frac{5}{9}(54) = 30$.

$C = \frac{5}{9}(F – 32)$; 86°F = 30°C

Exercise 17 – Taxi fare

Swap: $m = 3 + 2.5c$, so $c = \frac{m – 3}{2.5}$. In words: miles $= \frac{\text{fare} – 3}{2.5}$.

For $18: $\frac{18 – 3}{2.5} = 6$.

6 miles

Exercise 18 – Order matters

Discount first: $f(60) = 48$, then $g(48) = 43$.

Coupon first: $g(60) = 55$, then $f(55) = 44$.

Discount first: $43 beats $44

Exercise 19 – Growing circle

$A(r(t)) = \pi(3t)^{2} = 9\pi t^{2}$.

$A(2) = 9\pi \cdot 4 = 36\pi \approx 113.1$.

$A(t) = 9\pi t^{2}$; about 113.1 square inches

Exercise 20 – Find the error

$f(3) = 7$, so an inverse must send 7 back to 3. His function gives $\frac{1}{15}$.

Correct: $x = 2y + 1$, so $y = \frac{x – 1}{2}$; $\frac{7 – 1}{2} = 3$.

$f^{-1}(x) = \frac{x – 1}{2}$

Exercise 21 – Find the outer function

Let $u = x + 2$, so $x = u – 2$. Then $f(u) = (u – 2)^{2} + 4(u – 2) + 7 = u^{2} – 4u + 4 + 4u – 8 + 7 = u^{2} + 3$.

Check: $(x + 2)^{2} + 3 = x^{2} + 4x + 7$.

$f(x) = x^{2} + 3$

Key facts behind this composition inverses 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 answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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