Inverse Functions Worksheet Answer Key – Algebra 2

Problem 1 Adding, subtracting, and multiplying functions

(h + k)(x) = 5x + 2 + x² − 6 = x² + 5x − 4.

(h − k)(x) = 5x + 2 − x² + 6 = −x² + 5x + 8.

(h · k)(x) = (5x + 2)(x² − 6) = 5x³ + 2x² − 30x − 12.

Problem 2 Evaluating combined functions

h(3) = 17 and k(3) = 3, so (h + k)(3) = 20.

h(−2) = −8 and k(−2) = −2, so (h · k)(−2) = 16.

20 and 16

Problem 3 Composition at a number

k(5) = 11, so h(k(5)) = h(11) = 37.

h(5) = 13, so k(h(5)) = k(13) = 19.

37 and 19

Problem 4 Composition as an expression

h(k(x)) = (x − 5)² + 3 = x² − 10x + 25 + 3 = x² − 10x + 28.

k(h(x)) = (x² + 3) − 5 = x² − 2.

Problem 5 A function machine

k(8) = 24, then h(24) = 24 − 4 = 20.

20

Problem 6 A linear inverse

y = 4x + 9. Swap: x = 4y + 9. Solve: 4y = x − 9, so y = x − 94.

h⁻¹(x) = x − 94

Problem 7 An inverse with a fraction

y = 5x − 24. Swap: x = 5y − 24. Then 4x = 5y − 2, so y = 4x + 25.

h⁻¹(x) = 4x + 25

Problem 8 Verifying inverses

h(k(x)) = 6 · x − 16 + 1 = x − 1 + 1 = x.

k(h(x)) = (6x + 1) − 16 = 6x6 = x.

Both compositions equal x, so h and k are inverses.

Problem 9 Reflecting a graph

Swap the coordinates: (−5, −1), (−2, 0), (1, 1), (4, 2).

Equation: swap x = 3y − 2, so y = x + 23.

(−5, −1), (−2, 0), (1, 1), (4, 2); h⁻¹(x) = x + 23

Problem 10 A cube root inverse

y = x³ + 6. Swap: x = y³ + 6, so y³ = x − 6 and y = ∛(x − 6).

h⁻¹(14) = ∛8 = 2.

h⁻¹(x) = ∛(x − 6); h⁻¹(14) = 2

Problem 11 A restricted domain

y = x² − 7. Swap: x = y² − 7, so y² = x + 7 and y = √(x + 7) (positive root because the original inputs are x ≥ 0).

The outputs of h are y ≥ −7, so the domain of h⁻¹ is x ≥ −7.

h⁻¹(9) = √16 = 4 and h(4) = 16 − 7 = 9 ✓.

h⁻¹(x) = √(x + 7), x ≥ −7

Problem 12 Inverse of a square root function

y = √(x + 3), with x ≥ −3 and y ≥ 0. Swap: x = √(y + 3), so x² = y + 3 and y = x² − 3.

The outputs of h are y ≥ 0, so the inverse has domain x ≥ 0.

h⁻¹(x) = x² − 3, x ≥ 0

Problem 13 Domain of a composition

h(k(x)) = √(x + 8), which needs x + 8 ≥ 0: domain x ≥ −8.

k(h(x)) = √x + 8, which needs x ≥ 0: domain x ≥ 0.

√(x + 8), x ≥ −8; √x + 8, x ≥ 0

Problem 14 Inverse of a table

Swap each pair: (6, 1), (9, 2), (12, 3). So h⁻¹(12) = 3.

(6, 1), (9, 2), (12, 3); h⁻¹(12) = 3

Problem 15 A rational inverse

y = 5x + 3x − 2. Swap: x = 5y + 3y − 2, so xy − 2x = 5y + 3 and y(x − 5) = 2x + 3.

h⁻¹(x) = 2x + 3x − 5.

Check: h(h⁻¹(x)) = 5(2x + 3)/(x − 5) + 3(2x + 3)/(x − 5) − 2 = (10x + 15) + 3(x − 5)(2x + 3) − 2(x − 5) = 13x13 = x ✓.

h⁻¹(x) = 2x + 3x − 5

Problem 16 Converting temperatures

y = 95x + 32. Swap: x = 95y + 32, so y = 59(x − 32).

F⁻¹(98.6) = 59(66.6) = 37.

Check: F(F⁻¹(x)) = 95 · 59(x − 32) + 32 = x ✓.

C(h) = 59(h − 32); 98.6°F = 37°C

Problem 17 Order of discounts

Coupon first: C(60) = 50, then D(50) = 40.

Sale first: D(60) = 48, then C(48) = 38.

In general D(C(p)) = 0.8p − 8 and C(D(p)) = 0.8p − 10, so the second order is always $2 cheaper.

$40 and $38; apply the sale first

Problem 18 Taxi fare and miles

y = 2.5x + 3.5. Swap: x = 2.5y + 3.5, so y = x − 3.52.5.

For $21: 21 − 3.52.5 = 7.

m = C − 3.52.5; 7 miles

Problem 19 Making a function invertible

Complete the square: h(x) = (x − 3)² − 4. On x ≥ 3 the outputs are y ≥ −4.

Swap: x = (y − 3)² − 4, so (y − 3)² = x + 4 and y − 3 = √(x + 4) (the positive root, since y ≥ 3).

h⁻¹(x) = 3 + √(x + 4); domain x ≥ −4; range y ≥ 3

Problem 20 When do two compositions agree?

h(k(x)) = 4(x² + 3) − 1 = 4x² + 11.

k(h(x)) = (4x − 1)² + 3 = 16x² − 8x + 4.

Set equal: 4x² + 11 = 16x² − 8x + 4, so 12x² − 8x − 7 = 0.

x = 8 ± √(64 + 336)24 = 8 ± 2024, so x = 76 or x = –12.

x = 76 or x = –12