Function Notation, Domain, and Range — ACT Math

Introduction

Students who understand function notation can decode any f(g(x)) problem in seconds — students who don't waste 3 minutes on a medium-level question.

Function questions appear 4-6 times per ACT, split between medium and hard positions. They are disproportionately missed because students misread the notation.

By the end of this lesson you will be able to:

Evaluate functions using function notation, including composed functions f(g(x))
Determine the domain and range of functions defined by equations or graphs
Identify transformations that shift, reflect, or scale a function's graph

You'll find the domain of f(x) = √(2x − 8) / (x − 7) and explain why two different restrictions apply.

The Concept

The Core Rule

f(x) means 'the function f applied to input x.' f(3) means replace every x with 3. Composition f(g(x)) means apply g first, then apply f to the result. Domain = all valid inputs (x-values). Range = all possible outputs (f(x) values).

How the ACT tests this

Asking for f(g(a)) where you must apply g first, not f
Finding the domain of a function involving a square root or denominator
Using a graph to identify domain, range, or whether a graph represents a function

Function Notation and Evaluation

f(x) is just a named rule. f(2x + 1) means substitute (2x + 1) everywhere you see x in the formula. f(g(x)): first evaluate g(x) for the given input, then substitute that result into f.

f(a + b) ≠ f(a) + f(b) in general — this is a common misconception
Composition order matters: f(g(x)) ≠ g(f(x)) in general
Inverse function f⁻¹: swaps input and output; f(f⁻¹(x)) = x

Domain Restrictions

Two automatic restrictions: (1) No division by zero — set denominator ≠ 0. (2) No square root of a negative — set radicand ≥ 0. If both apply, apply both and intersect the results.

Denominator = 0: solve and exclude those x-values from the domain
Even root: radicand ≥ 0 (cube roots are unrestricted)
Stated domain: some problems give the domain explicitly — read carefully

Transformations

Starting from y = f(x): vertical shift up k → y = f(x) + k. Horizontal shift right h → y = f(x − h). Vertical stretch by a → y = a·f(x). Reflection over x-axis → y = −f(x). Reflection over y-axis → y = f(−x).

f(x − h): the sign inside is opposite to the direction of shift (right, not left)
y = f(2x) compresses the graph horizontally by factor 2
Vertical and horizontal transformations are independent of each other

Your strategy

For f(g(x)), circle g(x) and evaluate it first, writing down the number before touching f
For domain problems, list all restrictions, solve each, then combine with AND logic
For transformations, identify inside-the-function changes (horizontal) vs outside (vertical)
Picking numbers: test a simple value like x=0 to verify your domain restriction

Worked Examples

Easy Example 1 Adding 4 Instead Of Subtracting, Or Forgetting The Subtraction Entirely

If f(x) = 3x − 4, what is f(5)?

A. 9
B. 11 (Correct answer)
C. 13
D. 15
E. 19

Step 1

Substitute x = 5 into f(x) = 3x − 4

Step 2

f(5) = 3(5) − 4 = 15 − 4 = 11

Step 3

Answer: 11

Correct answer: B

Why B is correct

Correct: 15 − 4 = 11

Why other options are wrong

A: 3(5) − 6 = 9; subtracted wrong constant

C: 3(5) − 2 = 13; wrong constant

D: 3(5) = 15; forgot to subtract 4

E: 3(5) + 4 = 19; added instead of subtracted

⚠ Trap: Adding 4 instead of subtracting, or forgetting the subtraction entirely

Medium Example 2 Evaluating F First Instead Of G (reversing The Composition Order)

If f(x) = x² + 1 and g(x) = 2x − 3, what is f(g(2))?

A. 2 (Correct answer)
B. 3
C. 5
D. 6
E. 10

Step 1

First evaluate g(2): g(2) = 2(2) − 3 = 4 − 3 = 1

Step 2

Then evaluate f(1): f(1) = 1² + 1 = 2

Step 3

f(g(2)) = 2

Correct answer: A

Why A is correct

Correct: g(2)=1, f(1)=2

Why other options are wrong

B: Arithmetic error in second step

C: f(2) = 4+1=5; applied f directly to 2 instead of to g(2)

D: Arithmetic error

E: g(f(2)): f(2)=5, g(5)=2(5)−3=7; neither matches 10

⚠ Trap: Evaluating f first instead of g (reversing the composition order)

Hard Example 3 Forgetting That √0 = 0 Is Valid, So The Domain Uses ≥ Not >

What is the domain of f(x) = √(3x − 9) / (x − 5)?

A. x ≥ 3
B. x > 3
C. x ≥ 3 and x ≠ 5 (Correct answer)
D. x > 3 and x ≠ 5
E. All real numbers except x = 5

Step 1

Restriction 1 (square root): 3x − 9 ≥ 0 → 3x ≥ 9 → x ≥ 3

Step 2

Restriction 2 (denominator): x − 5 ≠ 0 → x ≠ 5

Step 3

Combine: x ≥ 3 AND x ≠ 5

Correct answer: C

Why C is correct

Correct: x ≥ 3 and x ≠ 5

Why other options are wrong

A: Ignores the denominator restriction x ≠ 5

B: Uses strict inequality for square root; x=3 gives √0 = 0, which is valid

D: Excludes x=3 incorrectly (√0 is defined and equals 0)

E: Ignores the square root restriction entirely

⚠ Trap: Forgetting that √0 = 0 is valid, so the domain uses ≥ not >

Strategy Tips

Write out each step of a composition separately on scratch paper
For domain, check for square roots AND denominators in the same function
Remember: √0 is defined and equals 0, so use ≥ not > for even-root domains
Transformations: changes INSIDE f(x − h) move the graph OPPOSITE to what you expect
Use the vertical line test on a graph to confirm a relation is a function

Common pitfalls

Reversing the order of composition: f(g(x)) ≠ g(f(x))

Using > instead of ≥ for square root domain restrictions

Missing a denominator restriction when a fraction is inside a square root

Function evaluation: 30-45 seconds. Composition: 60 seconds. Domain restrictions: 60-90 seconds. Write each step of a composition on scratch paper to avoid errors.

Summary

f(g(x)): always evaluate the inner function g first, then apply f
Domain: set radicands ≥ 0 and denominators ≠ 0, then intersect the conditions
Horizontal transformations inside f(x) move opposite to the sign

Find the domain of g(x) = √(x + 4) / (x² − x − 6). Identify both restrictions, then intersect them. Write the domain in interval notation.

Next: Linear, Quadratic, and Exponential Models All ACT Math lessons