ACT Math: Functions
Functions is a core Functions-domain topic on the ACT, and it underlies nearly every other function-related question you'll see. This free, complete lesson covers function notation and evaluating functions, domain and range, composite and inverse functions, and piecewise functions with even and odd functions. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.
1. Function Notation and Evaluating Functions
Function notation, f(x), names a rule that produces one output for each input. To evaluate f at a specific value, substitute that value everywhere x appears.
Worked Example: If f(x) = 2x² − 3x + 5, find f(−2).
f(−2) = 2(−2)² − 3(−2) + 5 = 2(4) + 6 + 5 = 8 + 6 + 5 = 19
1. If f(x) = 3x + 7, find f(4).
Explanation: f(4) = 3(4) + 7 = 12 + 7 = 19.
2. If f(x) = x² − 5x + 6, find f(3).
Explanation: f(3) = 3² − 5(3) + 6 = 9 − 15 + 6 = 0.
3. If g(x) = 2x² + x − 4, find g(−3).
Explanation: g(−3) = 2(9) + (−3) − 4 = 18 − 3 − 4 = 11.
4. If f(x) = 4x − 9, and f(a) = 15, what is a?
Explanation: 4a − 9 = 15, so 4a = 24, giving a = 6.
5. If f(x) = x² + 2x, find f(x + 1) in terms of x.
Explanation: Substitute (x + 1) for x: (x + 1)² + 2(x + 1) = x² + 2x + 1 + 2x + 2 = x² + 4x + 3.
2. Domain and Range
The domain is every valid input (x-value); the range is every output (y-value) the function actually produces. Two common domain restrictions: a denominator can never equal zero, and the expression under a square root can never be negative.
Worked Example: Find the domain of f(x) = 1/(x − 3).
The denominator cannot equal 0: x − 3 ≠ 0, so x ≠ 3.
Domain: all real numbers except 3.
1. Find the domain of f(x) = 1/(x + 5).
Explanation: x + 5 cannot equal 0, so x ≠ −5.
2. Find the domain of f(x) = √(x − 4).
Explanation: The expression under the square root must be non-negative: x − 4 ≥ 0, so x ≥ 4.
3. Find the domain of f(x) = 3/(x² − 9).
Explanation: x² − 9 ≠ 0, so x² ≠ 9, giving x ≠ 3 and x ≠ −3.
4. What is the range of f(x) = x² + 2 for all real x?
Explanation: Since x² ≥ 0 for all x, the minimum value of x² + 2 is 2, occurring at x = 0. The parabola opens upward, so the range is y ≥ 2.
5. What is the range of f(x) = −2x² + 8?
Explanation: The negative leading coefficient means the parabola opens downward, with a maximum value of 8 at x = 0. Range: y ≤ 8.
3. Composite and Inverse Functions
A composite function, f(g(x)), applies g first, then feeds that result into f. An inverse function, f⁻¹(x), reverses f: swap x and y in the equation, then solve for the new y.
Worked Example: If f(x) = 2x + 3 and g(x) = x² − 1, find f(g(2)).
First: g(2) = 4 − 1 = 3.
Then: f(3) = 2(3) + 3 = 9
1. If f(x) = x + 5 and g(x) = 3x, find f(g(4)).
Explanation: g(4) = 3(4) = 12. f(12) = 12 + 5 = 17.
2. If f(x) = 2x − 1 and g(x) = x² + 4, find g(f(3)).
Explanation: f(3) = 2(3) − 1 = 5. g(5) = 25 + 4 = 29.
3. If f(x) = x² − 3 and g(x) = 2x, find f(g(x)) in terms of x.
Explanation: Substitute g(x) = 2x into f: f(2x) = (2x)² − 3 = 4x² − 3.
4. If f(x) = 3x + 6, find the inverse function f⁻¹(x).
Explanation: Let y = 3x + 6. Swap x and y: x = 3y + 6. Solve for y: 3y = x − 6, so y = (x − 6)/3.
5. If f(x) = (x − 4)/2, find the inverse function f⁻¹(x).
Explanation: Let y = (x − 4)/2. Swap x and y: x = (y − 4)/2. Solve for y: 2x = y − 4, so y = 2x + 4.
4. Piecewise Functions and Even/Odd Functions
A piecewise function uses a different rule depending on which interval x falls into. A function is even if f(−x) = f(x) for all x (symmetric about the y-axis); it's odd if f(−x) = −f(x) for all x (symmetric about the origin).
Worked Example: f(x) = x + 3 if x < 0, and f(x) = x² − 1 if x ≥ 0. Find f(−2) and f(3).
Since −2 < 0, use the first rule: f(−2) = −2 + 3 = 1.
Since 3 ≥ 0, use the second rule: f(3) = 3² − 1 = 8.
1. Given f(x) = 2x if x < 1, and f(x) = x + 4 if x ≥ 1, find f(0).
Explanation: Since 0 < 1, use the first rule: f(0) = 2(0) = 0.
2. Using the same piecewise function, find f(1).
Explanation: Since 1 ≥ 1, use the second rule: f(1) = 1 + 4 = 5.
3. Using the same piecewise function, find f(−3).
Explanation: Since −3 < 1, use the first rule: f(−3) = 2(−3) = −6.
4. Is f(x) = x² an even function, an odd function, or neither?
Explanation: f(−x) = (−x)² = x² = f(x), so the function is even.
5. Is f(x) = x³ an even function, an odd function, or neither?
Explanation: f(−x) = (−x)³ = −x³ = −f(x), so the function is odd.
Common Mistakes to Avoid
- Confusing f(x + 1) with f(x) + 1. The first substitutes x + 1 everywhere x appears in the formula; the second just adds 1 to the final output. These give very different results for anything beyond a purely linear function.
- Evaluating a composite function in the wrong order. In f(g(x)), g is applied first, then f is applied to that result. f(g(2)) and g(f(2)) are generally not the same value.
- Forgetting to swap x and y when finding an inverse. Finding f⁻¹(x) always starts with swapping the variables, then solving for the new y, skipping the swap gives back the original function instead.
- Forgetting a domain restriction from a denominator or a square root. Always check for values that would make a denominator zero or a square root negative before stating the full domain.
- Applying the wrong piece of a piecewise function. Carefully check which inequality the input value satisfies before choosing which sub-formula to use, this is where most piecewise errors happen.
Frequently Asked Questions
What's the difference between the domain and the range of a function?
The domain is the set of valid inputs (x-values) you're allowed to plug into the function. The range is the set of outputs (y-values) that actually come out the other end.
Why does order matter for composite functions?
f(g(x)) means "first apply g to x, then apply f to that result," while g(f(x)) reverses that order. Since most functions transform their input differently, applying them in a different order generally produces a different final answer.
Is there a shortcut for checking if a function is even or odd?
Substitute −x for x throughout the function and simplify. If you get back exactly the original function, it's even. If you get back the exact negative of the original function, it's odd. If neither happens, the function is neither even nor odd.
Where can I practice more problems like these?
The Functions quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.