Inverse Functions Test Answer Key with Scoring – Algebra 2

Question 1 Combining functions (2 points)

(h + k)(x) = x² − 3x + 2x + 7 = x² − x + 7.

h(−2) = 4 + 6 = 10 and k(−2) = 3, so (h · k)(−2) = 30.

Scoring: 1 point each.

Question 2 Composition (3 points)

h(k(x)) = 3(x² + 1) − 2 = 3x² + 1.

k(h(x)) = (3x − 2)² + 1 = 9x² − 12x + 5.

k(2) = 5 and h(5) = 13, so h(k(2)) = 13.

Scoring: 1 point each.

Question 3 Find an inverse (3 points)

y = 3x − 85. Swap: x = 3y − 85, so 5x = 3y − 8 and y = 5x + 83.

h⁻¹(x) = 5x + 83

Check: h(h⁻¹(x)) = 3 · [[5x + 83 − 8|5]] = 5x + 8 − 85 = x ✓.

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

Question 4 Inverse from a graph (3 points)

a) Swap coordinates: (1, 0), (2, 2), (3, 4).

b) h(x) = x2 + 1. Swap: x = y2 + 1, so y = 2x − 2. h⁻¹(x) = 2x − 2

c) The point on the line y = x: (2, 2).

Scoring: 1 point per part.

Question 5 Restricted domain (3 points)

a) y = x² + 5. Swap: x = y² + 5, so y = √(x − 5) (positive root). h⁻¹(x) = √(x − 5)

b) The outputs of h are y ≥ 5, so the domain of h⁻¹ is x ≥ 5, and its range is y ≥ 0.

Scoring: 1 point for swapping, 1 for the positive root, 1 for the domain and range.

Question 6 Domain of a composition (2 points)

h(k(x)) = √(x² − 7), which requires x² ≥ 7: domain x ≤ −√7 or x ≥ √7.

k(h(x)) = (√(x − 7))² = x − 7, which requires x − 7 ≥ 0: domain x ≥ 7.

Scoring: 1 point each, including the domain.

Question 7 Sale and coupon (2 points)

Sale first: S(48) = 36, then K(36) = 28.

Coupon first: K(48) = 40, then S(40) = 30.

Sale first costs $28, which is $2 cheaper than $30

Scoring: 1 point for both prices, 1 point for the conclusion.

Question 8 Verify inverses (2 points)

h(k(x)) = (∛(x + 1))³ − 1 = x + 1 − 1 = x.

k(h(x)) = ∛((x³ − 1) + 1) = ∛(x³) = x.

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

Scoring: 1 point for each composition.