ACT Math study guide · Chapter 16
Inverse Functions
A function's inverse undoes what the original function does. To find an inverse algebraically, swap x and y in the equation, then solve for the new y. The domain of f⁻¹ equals the range of f, and the range of f⁻¹ equals the domain of f, since the inverse simply reverses the input-output relationship.
f(f⁻¹(x)) = x for every x in the domain of f⁻¹, by definition. A question that asks for f(f⁻¹(5)) or f⁻¹(f(5)) is testing whether this property is recognized directly, without needing to find either function's formula at all.
When a function like x² is not one-to-one over its full domain, restricting the domain (to x ≥ 0, for instance) makes an inverse possible. The restriction carries through to the inverse: the sign in front of a square root in the inverse formula is chosen to match the restricted piece the original function came from.
Where this appears as a model: converting a described relationship back the other direction. A Celsius-to-Fahrenheit conversion F(C)=(9/5)C+32 gives F(20)=68; its inverse, C(F)=(5/9)(F−32), recovers C(68)=20, confirming the two functions undo each other.
▸ Finding an inverse algebraically
Worked Example Find the inverse of f(x) = (2x + 3) ⁄ 5.
Swap x and y, then solve for the new y.
x = (2y + 3)/5
5x = 2y + 3
y = (5x − 3)/2
▸ Restricted domains and the inverse
Worked Example Let f(x) = x² − 5, restricted to x ≥ 0. Find f⁻¹(11).
The restriction to x ≥ 0 means the inverse takes the positive square root.
f⁻¹(x) = √(x + 5) on this restricted domain
f⁻¹(11) = √16 = 4
▸ Practice
1. Let f(x) = 4x + 9. Find f⁻¹(21).
A) 93 B) 3 C) 12 D) −3
2. Let f(x) = 5x − 2 and g(x) = (x + 2) ⁄ 5. Find g(f(7)).
F) 33 G) 7 H) 5 J) −3
3. Let f(x) = (x − 1) ⁄ 2. Find f⁻¹(3).
A) 1 B) 4 C) 7 D) −7
4. Let f(x) = x² − 4, restricted to x ≤ 0. Find f⁻¹(5).
F) 3 G) 9 H) −3 J) −1
5. A function f has domain [2, 10] and range [−3, 7]. What is the domain of f⁻¹?
A) [−3, 7] B) [2, 10] C) [−7, 3] D) [−10, −2]
6. Let f(x) = 2x³ − 1. Find f⁻¹(15).
F) 2 G) 8 H) 4 J) −2
7. Find the inverse of f(x) = 3x + 7.
A) (x + 7) ⁄ 3 B) 3x − 7 C) (x − 7) ⁄ 3 D) (x + 3) ⁄ 7
8. Let f(x) = 4x − 5 and g(x) = (x + 5) ⁄ 4. Find g(f(6)).
F) 19 G) 6 H) 24 J) −6
9. Let f(x) = x² + 3, restricted to x ≥ 0. Find f⁻¹(19).
A) 16 B) 4 C) 22 D) −4
10. If f has domain [2, 9] and range [−3, 7], find the domain of f⁻¹.
F) [−3, 7] G) [2, 9] H) [−9, −2] J) [−7, 3]
11. Find the inverse of f(x) = (x − 4) ⁄ 2.
A) 2x − 4 B) x/2 + 4 C) x/2 − 4 D) 2x + 4
12. Let f(x) = (x − 4) ⁄ 2. Find the value of x for which f(x) = 10.
F) 24 G) 3 H) 7 J) 10
13. Let f(x) = 3x³ + 1. Find f⁻¹(82).
A) 9 B) 3 C) 27 D) 4
14. Let f(x) = (x − 4) ⁄ 2. Find f(f⁻¹(5)).
F) 0 G) 5 H) 10 J) −5
15. Find the inverse of f(x) = 6x − 11.
A) (x + 11) ⁄ 6 B) 6x + 11 C) (x − 11) ⁄ 6 D) 6x + 6
16. Let f(x) = 7x − 3 and g(x) = (x + 3) ⁄ 7. Find g(f(4)).
F) 25 G) −4 H) 4 J) 28
17. Let f(x) = x² + 7, restricted to x ≤ 0. Find f⁻¹(23).
A) −4 B) 4 C) 16 D) −16
18. A function f has domain [3, 12] and range [−5, 9]. Find the domain of f⁻¹.
F) [3, 12] G) [−12, −3] H) [−9, 5] J) [−5, 9]
19. Find the inverse of f(x) = (x + 6) ⁄ 3.
A) 3x + 6 B) x/3 − 6 C) x/3 + 6 D) 3x − 6
20. Let f(x) = 4x³ − 3. Find f⁻¹(29).
F) 8 G) 4 H) −2 J) 2
▸ Answers and Explanations
1: B. f⁻¹(x) = (x − 9) ⁄ 4, so f⁻¹(21) = 12 ⁄ 4 = 3. Choice A applies the original function's operations to 21 instead of the inverse, computing 4(21)+9=93.
2: G. f(7) = 5(7) − 2 = 33, then g(33) = (33 + 2) ⁄ 5 = 7. Since f and g are genuine inverses, this composition returns the original input, 7. Choice F reports f(7) alone instead of completing the composition by evaluating g at that value.
3: C. f⁻¹(x) = 2x + 1, so f⁻¹(3) = 2(3) + 1 = 7. Choice D drops the sign while solving for the inverse, computing 2(3)−1 instead of 2(3)+1.
4: H. f⁻¹(x) = −√(x + 4) on this restricted domain, so f⁻¹(5) = −√9 = −3. Choice F drops the negative sign required by the restricted domain of the inverse.
5: A. The domain of f⁻¹ equals the range of f, which is [−3, 7]. Choice B reports the domain of f itself instead of the domain of its inverse.
6: F. Solving 2x³ − 1 = 15 gives x³ = 8, so x = 2. Choice G reports the value of x³ instead of completing the cube root to solve for x.
7: C. Swap and solve: x=3y+7, so y=(x−7)/3. Choice A drops the sign while solving for the inverse, adding 7 instead of subtracting it.
8: G. f(6)=4(6)−5=19; g(19)=(19+5)/4=6. Choice F reports f(6) alone instead of completing the composition by evaluating g at that value.
9: B. x²+3=19 gives x²=16; since x≥0, x=4. Choice A reports the value of x² instead of completing the square root to solve for x.
10: F. The domain of f⁻¹ equals the range of f: [−3, 7]. Choice G reports the domain of f itself instead of the domain of its inverse.
11: D. Swap and solve: x=(y−4)/2, so 2x=y−4, giving y=2x+4. Choice F drops the sign while solving for the inverse, subtracting 4 instead of adding it.
12: F. (x−4)/2=10 gives x−4=20, so x=24. Choice G divides by 2 before adding 4, reversing the correct order of operations.
13: B. 3x³+1=82 gives 3x³=81, so x³=27, giving x=3. Choice A divides 27 by 3 instead of taking its cube root.
14: G. By the definition of an inverse function, f(f⁻¹(x))=x for any x in the domain, so f(f⁻¹(5))=5. Choice F misapplies the inverse property, computing f(x)−f⁻¹(x) instead of recognizing that f(f⁻¹(x))=x directly.
15: A. Swap and solve: x=6y−11, so y=(x+11)/6. Choice C drops the sign while solving for the inverse, subtracting 11 instead of adding it.
16: H. f(4)=7(4)−3=25; g(25)=(25+3)/7=4. Choice F reports f(4) alone instead of completing the composition by evaluating g at that value.
17: A. x²+7=23 gives x²=16; since x≤0, x=−4. Choice B drops the negative sign required by the restricted domain of the inverse.
18: J. The domain of f⁻¹ equals the range of f: [−5, 9]. Choice F reports the domain of f itself instead of the domain of its inverse, which equals the range of f.
19: D. Swap and solve: x=(y+6)/3, so 3x=y+6, giving y=3x−6. Choice A drops the sign while solving for the inverse, adding 6 instead of subtracting it.
20: J. 4x³−3=29 gives 4x³=32, so x³=8, giving x=2. Choice F reports the value of x³ instead of completing the cube root to solve for x.