27 exam-style questions on Composite functions in context; inverse with domain restriction — topic 2.7 of Functions in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.
Given f(x) = 3x − 2 and g(x) = x + 5, calculate (f ∘ g)(4).
Answer: 25
g(4) = 4 + 5 = 9, then f(9) = 3(9) − 2 = 27 − 2 = 25.
Given f(x) = x² and g(x) = 2x + 1, calculate (g ∘ f)(3).
Answer: 19
f(3) = 3² = 9, then g(9) = 2(9) + 1 = 18 + 1 = 19. (Note (f ∘ g)(3) would give 49, a common mix-up.)
Given f(x) = 4x + 3, determine the inverse function f⁻¹(x).
Answer: (x − 3)/4
Let y = 4x + 3, swap and solve: x = 4y + 3 → y = (x − 3)/4. Adding 3 instead of subtracting is a common error.
A currency converter uses f(x) = 0.85x (dollars to euros). Determine the inverse function that converts euros back to dollars.
Answer: f⁻¹(x) = x/0.85
y = 0.85x, so x = y/0.85; the inverse is f⁻¹(x) = x/0.85 (divide by the rate to reverse the conversion).
For f(x) = x² with domain x ≥ 0, determine the inverse function f⁻¹(x).
Answer: √x
Restricting to x ≥ 0 makes f one-to-one, so the inverse is f⁻¹(x) = √x (the positive root only).
Given f(x) = 2x and g(x) = x − 3, determine an expression for (f ∘ g)(x).
Answer: 2x − 6
(f ∘ g)(x) = f(x − 3) = 2(x − 3) = 2x − 6. Forgetting to multiply the −3 by 2 gives the distractor 2x − 3.
That was 6 of 27 Composite functions in context; inverse with domain restriction questions.
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