Inverse Functions Worksheet Answer Key – Precalculus

Problem 1 Evaluate with Function Notation

f(−3) = 2(9) − 5(−3) + 7 = 18 + 15 + 7 = 40.

f(4) = 2(16) − 20 + 7 = 32 − 20 + 7 = 19.

Problem 2 Domain Using Two Restrictions

(a) Need 4x + 9 ≥ 0, so x ≥ −9/4. Domain: [−9/4, ∞).

(b) Need x + 6 ≠ 0, so x ≠ −6. Domain: (−∞, −6) ∪ (−6, ∞).

Problem 3 Domain and Range from a Graph

The curve starts at x = −3 (closed) and stops before x = 5 (open): domain [−3, 5).

The highest output is 4. The lowest output is 0, reached at (−3, 0), so the open dot does not matter: range [0, 4].

Problem 4 Evaluate Two Compositions

g(2) = 9, so f(9) = 81 − 3 = 78.

f(2) = 1, so g(1) = 5.

Problem 5 Build Composite Functions

(f ∘ g)(x) = 6(x² − 4) + 1 = 6x² − 23.

(g ∘ f)(x) = (6x + 1)² − 4 = 36x² + 12x + 1 − 4 = 36x² + 12x − 3.

Problem 6 Inverse of a Linear Function

Swap: x = 54y + 3. Subtract 3: x − 3 = 54y. Multiply by 45: y = 45(x − 3).

f⁻¹(x) = 4x − 125.

Problem 7 Reading Inverse Values from a Table

f(5) = 15, so f⁻¹(15) = 5. f(11) = 30, so f⁻¹(30) = 11.

f(f⁻¹(22)) = 22 because a function and its inverse cancel.

The output 10 does not appear in the table, so f⁻¹(10) cannot be found from these data.

Problem 8 Inverse Values from a Graph

Swap the coordinates: (4, 3), (0, 1) and (−4, −1) lie on the inverse.

f⁻¹(4) = 3, f⁻¹(0) = 1, f⁻¹(−4) = −1.

Problem 9 Domain of a Root over a Binomial

The root requires x + 5 ≥ 0, so x ≥ −5. The denominator requires x ≠ 2, and 2 lies inside that range.

Domain: [−5, 2) ∪ (2, ∞).

Problem 10 Verify Inverse Functions

f(g(x)) = 2 · x − 12 + 1 = x − 1 + 1 = x.

g(f(x)) = ∛(2x³2) = ∛(x³) = x.

Both compositions equal x, so they are inverses.

Problem 11 Decompose a Composite Function

(a) Inner g(x) = 5x − 2, outer f(u) = u⁴: h = f ∘ g.

(b) Inner g(x) = x² + 9, outer f(u) = 1/u: k = f ∘ g.

Problem 12 Restrict, Then Invert

On x ≥ −3 the graph rises, so it is one-to-one. Swap: x = (y + 3)² − 4, so y + 3 = √(x + 4) (positive root because y ≥ −3).

f⁻¹(x) = √(x + 4) − 3, domain [−4, ∞), range [−3, ∞).

Problem 13 Inverse of a Rational Function

Swap: x = y + 1y − 4. Then xy − 4x = y + 1, so y(x − 1) = 4x + 1.

f⁻¹(x) = 4x + 1x − 1, and x = 1 is excluded (1 is not in the range of f).

Problem 14 Domain of a Composite

(f ∘ g)(x) = √(x² − 10 − 1) = √(x² − 11).

Need x² ≥ 11, so |x| ≥ √11. Domain: (−∞, −√11] ∪ [√11, ∞).

Problem 15 Horizontal Line Test and Restriction

(a) x³ − 3x = x(x² − 3) = 0 gives x = 0, √3, −√3. So g(0) = g(√3) = g(−√3) = 0, and g is not one-to-one.

(b) The turning points are at x = ±1, so g is one-to-one on [1, ∞) (also on (−∞, −1] or [−1, 1]).

Problem 16 Taxi Fare and Its Inverse

Swap and solve: m = (C − 2.50) ÷ 1.75, that is f⁻¹(x) = x − 2.501.75. It converts a fare in dollars to miles driven.

f⁻¹(23.50) = 21 ÷ 1.75 = 12 miles.

Problem 17 Lemonade Stand Revenue

r(c(t)) = 1.75(12t + 8) − 20 = 21t + 14 − 20 = 21t − 6.

At t = 5: 105 − 6 = $99.

Solve 21t − 6 = 162: t = 168 ÷ 21 = 8 hours.

Problem 18 Spreading Ripple Area

A(r(t)) = π(4 + 3t)².

At t = 2: π(10)² = 100π ≈ 314.16 ft².

(4 + 3t)² = 169 gives 4 + 3t = 13 (radius is positive), so t = 3 seconds.

Problem 19 Cabin Cost Split

From s = 480n + 4 we get n + 4 = 480s, so n = 480s − 4. Using x as the input: f⁻¹(x) = 480x − 4, the number of students for a share of x dollars.

f⁻¹(32) = 15 − 4 = 11 students.

Problem 20 Restrict, Invert and Evaluate

h(x) = 2(x² − 1) + 3 = 2x² + 1. On x ≥ 0 it is one-to-one.

Swap: x = 2y² + 1, so y = √(x − 12) (positive root since y ≥ 0). Hence h⁻¹(x) = √(x − 12).

h⁻¹(19) = √9 = 3. Check: h(3) = 19.