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ACT Math Functions: 18 Practice Problems with Step-by-Step Explanations

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ACT Math Functions: 18 Practice Problems with Step-by-Step Explanations

Functions are one of the broadest topics on the ACT Math section, covering everything from simply plugging a number into f(x) to composing two functions together, finding inverse functions, and reading the domain or range directly off a graph. On a typical ACT you'll see functions questions scattered across the whole test, from early, single-step evaluation problems to some of the hardest problems near the end of the section, where composition, restricted domains, and rational functions with asymptotes all get combined into one question.

A few core skills come up again and again: evaluating f(x) at a given input by direct substitution; composing functions, where (f ∘ g)(x) means "plug g(x) into f," always working from the inside out; finding an inverse function by swapping x and y and solving for the new y; and determining domain and range, which usually means checking for values that make a denominator zero or a value under a square root negative. Graph-based questions test the same ideas visually — the range of a function is simply the set of y-values the graph actually reaches, and a rational function's domain restrictions show up as vertical asymptotes or open circles ("holes") on its graph.

Common traps include composing functions in the wrong order (f(g(x)) is not the same as g(f(x)) unless the problem says otherwise), forgetting that a "hole" in a graph still makes that x-value excluded from the domain even though the function looks defined nearby, and mixing up when a rational or square-root expression requires the denominator to be nonzero versus the radicand to be non-negative. Work through the 18 problems below — including three that reference a graph — then click to reveal each step-by-step explanation.

On this page:

  1. Problem 1 — Reading a graph (range)
  2. Problem 2 — Inverse function
  3. Problem 3 — Evaluating a function
  4. Problem 4 — Adding two functions
  5. Problem 5 — Composing functions
  6. Problem 6 — Composing functions with fractions
  7. Problem 7 — Composing functions
  8. Problem 8 — Range from a graph
  9. Problem 9 — Domain of a rational function
  10. Problem 10 — Range on a restricted domain
  11. Problem 11 — Domain from a graph
  12. Problem 12 — Function on a restricted domain
  13. Problem 13 — Domain of a composed function
  14. Problem 14 — Multiplying functions
  15. Problem 15 — Transforming a function
  16. Problem 16 — Domain of a rational function
  17. Problem 17 — Solving a composed equation
  18. Problem 18 — Solving for a constant in a composition

Practice Problems

Problem 1

The fluctuation of the water depth at a wetland is shown in the figure below. One of the values gives the positive difference, in inches, between the greatest water depth and least water depth shown in the graph. Which value is it?

Depth (inches) vs. number of hours after 1pm

  • A) 2
  • B) 3
  • C) 4
  • D) 6
Correct Answer: D

Reading the graph, the greatest depth shown is about 7 inches, and the least depth shown is about 1 inch.

The positive difference between them is 7 − 1 = 6 inches.

Problem 2

For the function f(x) = (x+1)/3, what is the value of f−1(x)?

  • A) (3x−1)/3
  • B) 3x − 1
  • C) (x+1)/3
  • D) (−x−1)/3
Correct Answer: B

Start by writing the function as y = (x+1)/3. To find the inverse, swap x and y: x = (y+1)/3.

Solve for y: multiply both sides by 3 to get 3x = y + 1, then subtract 1: y = 3x − 1. So f−1(x) = 3x − 1.

Problem 3

Given f(x) = −6x3 + 2x2, what is f(−3) equal to?

  • A) −180
  • B) −144
  • C) 144
  • D) 180
Correct Answer: D

Substitute −3 for x: f(−3) = −6(−3)3 + 2(−3)2 = −6(−27) + 2(9).

= 162 + 18 = 180.

A common error is mishandling the sign of (−3)3 = −27 — remember an odd power keeps the negative sign, while an even power makes it positive.

Problem 4

If f(x) = −3x2 − x and g(x) = −7x + 1, (f+g)(2) is equal to:

  • A) 1
  • B) −3
  • C) −10
  • D) −27
Correct Answer: D

(f+g)(2) means f(2) + g(2). First, f(2) = −3(2)2 − 2 = −12 − 2 = −14.

Next, g(2) = −7(2) + 1 = −14 + 1 = −13.

Add them: −14 + (−13) = −27.

Problem 5

For the functions f(x) = 3x + 7 and g(x) = 2x − 3, (f ∘ g)(5) is equal to:

  • A) 7
  • B) 11
  • C) 22
  • D) 28
Correct Answer: D

(f ∘ g)(5) means f(g(5)) — work from the inside out. First find g(5) = 2(5) − 3 = 10 − 3 = 7.

Then substitute that result into f: f(7) = 3(7) + 7 = 21 + 7 = 28.

Problem 6

If f(x) = x + 1/(2x) and g(x) = 1/x, what is the value of f(g(1/4))?

  • A) −6
  • B) −4
  • C) 4
  • D) 4 1/8
Correct Answer: D

Work from the inside out. First find g(1/4) = 1/(1/4) = 4.

Then substitute 4 into f: f(4) = 4 + 1/(2·4) = 4 + 1/8 = 4 1/8.

Problem 7

Functions f and g are defined as f(x) = −5x + 2 and g(x) = 10 − f(x). The value of g(3) is:

  • A) −13
  • B) −3
  • C) 7
  • D) 23
Correct Answer: D

Since g(x) is defined in terms of f(x), first find f(3): f(3) = −5(3) + 2 = −15 + 2 = −13.

Then substitute into g: g(3) = 10 − f(3) = 10 − (−13) = 10 + 13 = 23.

Problem 8

The graph of f(x) = 4cos(3x) + 3 is shown below. What is the range for this function?

f(x) = 4cos(3x) + 3

  • A) −1 < y < 7
  • B) −5 < y < 5
  • C) −1 ≤ y ≤ 7
  • D) −1 ≤ y ≤ 4
Correct Answer: C

Cosine always oscillates between −1 and 1, inclusive. Multiplying by 4 stretches that range to between −4 and 4, and adding 3 shifts it up to between −1 and 7.

Since cosine does reach its maximum and minimum values (−1 and 1 exactly), the range is inclusive: −1 ≤ y ≤ 7, not a strict inequality.

Problem 9

The domain of f(x) = (27−x)/(x2−6x+9) is all real numbers EXCEPT:

  • A) 27 only
  • B) 9 only
  • C) 3 only
  • D) −3 and 3 only
Correct Answer: C

A function is undefined wherever its denominator equals zero. Factor the denominator: x2 − 6x + 9 = (x − 3)2.

Setting (x−3)2 = 0 gives x = 3 as the only excluded value (it's a repeated root, not two different values), so the domain excludes 3 only.

Problem 10

For the function g defined by g(x) = 12 − (3/4)x with domain −16 ≤ x ≤ 8, what is the range of g?

  • A) −24 ≤ y ≤ 24
  • B) −16 ≤ y ≤ 8
  • C) 6 ≤ y ≤ 24
  • D) 0 ≤ y ≤ 6
Correct Answer: C

Since g is linear with a negative slope, it's decreasing, so the range endpoints come from evaluating g at the two domain endpoints. At x = −16: g(−16) = 12 − (3/4)(−16) = 12 + 12 = 24.

At x = 8: g(8) = 12 − (3/4)(8) = 12 − 6 = 6. Since the function decreases from x = −16 to x = 8, the range runs from 6 up to 24: 6 ≤ y ≤ 24.

Problem 11

The graph of f(x) = (−x2−x+12)/(5x2−25x+30) is shown below. The domain of f(x) is all values of x EXCEPT:

f(x) = (−x²−x+12)/(5x²−25x+30), with a vertical asymptote at x = 2 and a hole at x = 3

  • A) 2
  • B) 3
  • C) 0 and 2
  • D) 2 and 3
Correct Answer: D

Factor both numerator and denominator. The numerator −x2−x+12 factors as −(x+4)(x−3), and the denominator 5x2−25x+30 factors as 5(x−2)(x−3).

Setting the original denominator equal to 0 gives x = 2 and x = 3. Even though the (x−3) factor cancels with the numerator, creating a hole rather than an asymptote at x = 3, the function is still undefined at x = 3 in the original expression, so both x = 2 and x = 3 are excluded from the domain. This matches the graph, which shows a vertical asymptote at x = 2 and an open circle (hole) at x = 3.

Problem 12

A function is defined as f(z) = 3z − 13, and its domain is the set of integers from 1 − 20, including 1 and 20. For how many values of z is f(z) negative?

  • A) 0
  • B) 4
  • C) 5
  • D) 12
Correct Answer: B

Set up the inequality f(z) < 0: 3z − 13 < 0. Solving: 3z < 13, so z < 13/3 ≈ 4.33.

Since z must be an integer from 1 to 20, the values satisfying z < 4.33 are z = 1, 2, 3, and 4 — that's 4 values.

Problem 13

Consider the functions f(x) = √x and g(x) = 4x + 5. What is the domain of f(g(x))?

  • A) x ≥ −5/4
  • B) x > −5/4
  • C) x ≤ 4/5
  • D) x ≥ 5/4
Correct Answer: A

f(g(x)) = √(4x+5). A square root is only defined when the expression underneath it is non-negative, so set up the inequality: 4x + 5 ≥ 0.

Solving: 4x ≥ −5, so x ≥ −5/4. Since the boundary value itself is allowed (the square root of 0 is defined), the inequality is not strict.

Problem 14

If f(x) = 3x2 − 5x + 2 and g(x) = x2 + 3, which of the following expressions represents (fg)(x)?

  • A) 3x4 + 6x2 − 15
  • B) 3x4 + 11x2 − 6x − 9
  • C) 3x4 − 5x3 + 11x2 − 15x + 6
  • D) 9x4 − 30x3 + 37x2 − 20x + 4
Correct Answer: C

(fg)(x) means f(x) multiplied by g(x): (3x2 − 5x + 2)(x2 + 3). Distribute each term of the first polynomial across the second.

3x2(x2+3) = 3x4 + 9x2; −5x(x2+3) = −5x3 − 15x; 2(x2+3) = 2x2 + 6.

Combine like terms: 3x4 − 5x3 + (9x2+2x2) − 15x + 6 = 3x4 − 5x3 + 11x2 − 15x + 6.

Problem 15

The function f is defined by f(x) = 3x2 − 5x + 10. If the function f(x−2) is written in the form ax2 + bx + c, what is the value of a + b + c?

  • A) 30
  • B) 18
  • C) 10
  • D) 8
Correct Answer: B

Substitute (x−2) for x in f: f(x−2) = 3(x−2)2 − 5(x−2) + 10. Expand (x−2)2 = x2 − 4x + 4, so 3(x−2)2 = 3x2 − 12x + 12.

Expand the rest: −5(x−2) = −5x + 10. Combine everything: 3x2 − 12x + 12 − 5x + 10 + 10 = 3x2 − 17x + 32.

So a = 3, b = −17, c = 32, and a + b + c = 3 − 17 + 32 = 18.

Problem 16

f(x) = ((x−2)4−3) / ((x−3)2−4(x−2)+8). For what value of x is the function above undefined?

  • A) 2
  • B) 3
  • C) 4
  • D) 5
Correct Answer: D

The function is undefined where the denominator equals 0. Simplify the denominator first: (x−3)2 − 4(x−2) + 8 = (x2−6x+9) − 4x + 8 + 8.

Combine like terms: x2 − 6x + 9 − 4x + 16 = x2 − 10x + 25, which factors as (x−5)2.

Setting (x−5)2 = 0 gives x = 5 as the only value that makes the denominator zero.

Problem 17

Given functions f(x) = x2 + 6x − 7 and g(x) = 3x, for what values of x does g(f(x)) = 1?

  • A) 1 only
  • B) −1 and 7 only
  • C) 0 and −1 only
  • D) −7 and 1 only
Correct Answer: D

g(f(x)) = 3f(x). Since any nonzero base raised to the power 0 equals 1, this equation is satisfied exactly when the exponent f(x) equals 0.

Solve f(x) = 0: x2 + 6x − 7 = 0 factors as (x+7)(x−1) = 0, giving x = −7 or x = 1.

Problem 18

The 2 functions f and g are defined as f(x) = 6x − a and g(x) = 4x + 4, where a is a real number. If f(g(x)) = g(f(x)), then what is the value of a?

  • A) 0
  • B) −20/3
  • C) 3/2
  • D) 28/5
Correct Answer: B

Find both compositions. f(g(x)) = 6(4x+4) − a = 24x + 24 − a. g(f(x)) = 4(6x−a) + 4 = 24x − 4a + 4.

Set them equal: 24x + 24 − a = 24x − 4a + 4. The 24x terms cancel from both sides, leaving 24 − a = −4a + 4.

Solve for a: add 4a to both sides and subtract 4: 20 = −3a, so a = −20/3.

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