Composite functions; inverse functions practice questions.

Let f(x) = x − 4 and g(x) = 3x. Determine an expression for (g ∘ f)(x).

• 3x − 4
• 3x − 12
• x − 12
• 3x + 12

Answer: 3x − 12

(g ∘ f)(x) means apply f first: g(x − 4) = 3(x − 4) = 3x − 12. The result 3x − 4 comes from composing in the wrong order, (f ∘ g)(x).

The function h(x) = √(x + 2) is written as a composite h = f ∘ g. Which pair of functions is correct?

• f(x) = x + 2 and g(x) = √x
• f(x) = √x and g(x) = x − 2
• f(x) = x − 2 and g(x) = √x
• f(x) = √x and g(x) = x + 2

Answer: f(x) = √x and g(x) = x + 2

In f ∘ g the inner function g acts first: g(x) = x + 2 is applied, then the square root, giving f(g(x)) = √(x + 2). Choosing f(x) = x + 2, g(x) = √x would give √x + 2 instead.

Given f(x) = (x − 5)/2, find an expression for f⁻¹(x).

• 2x + 5
• 2x − 5
• (x + 5)/2
• 2/(x − 5)

Answer: 2x + 5

Set y = (x − 5)/2 and solve for x: x = 2y + 5, so f⁻¹(x) = 2x + 5. Note f⁻¹(x) is not the reciprocal 1/f(x) — a common trap.

Let f(x) = 1/x and g(x) = x − 3. State the largest possible domain of (f ∘ g)(x).

• x ≠ 0
• x ≠ −3
• x ≠ 3
• all real numbers

Answer: x ≠ 3

(f ∘ g)(x) = 1/(x − 3), so the denominator forbids x = 3. The restriction of f alone (x ≠ 0) no longer applies after composition.

Let f(x) = 2x − 1. Solve the equation (f ∘ f)(x) = 5.

• x = 2
• x = 3
• x = 1
• x = 3/2

Answer: x = 2

(f ∘ f)(x) = 2(2x − 1) − 1 = 4x − 3. Then 4x − 3 = 5 gives x = 2. Solving only f(x) = 5 gives the distractor x = 3.

Let f(x) = x² + 1 with domain x ≥ 0. Determine f⁻¹(x).

• √x − 1
• √(x − 1)
• (x − 1)²
• √(x + 1)

Answer: √(x − 1)

From y = x² + 1, x = √(y − 1) since x ≥ 0, so f⁻¹(x) = √(x − 1). Subtract 1 before taking the root, not after.