These eight questions are original and ramp from substitution to unknown functions. Write your own working first, then open each answer.
Use the timed original practice session builder if you want to attempt them against the clock. The lessons for the whole chapter are in Functions.
Questions 1 to 4: notation, range and composites
Question 1. f(x) = 5 − 2x has domain {−1, 0, 2, 3}. Find the range.
Answer
f(−1) = 7, f(0) = 5, f(2) = 1, f(3) = −1. The range is {−1, 1, 5, 7}.
Question 2. g(x) = x² − 4 has domain −3 ≤ x ≤ 2. Find the range.
Answer
g(−3) = 5 and g(2) = 0. The rule turns at x = 0, where g(0) = −4, and 0 lies inside the domain. The smallest output is −4 and the largest is 5.
The range is −4 ≤ g(x) ≤ 5.
Question 3. h(x) = |2x − 6|. Find h(1), then solve h(x) = 4.
Answer
h(1) = |−4| = 4.
For h(x) = 4, either 2x − 6 = 4 or 2x − 6 = −4. So x = 5 or x = 1. Both inputs give 4, which is why h(1) = 4 matched.
Question 4. f(x) = x + 5 and g(x) = x² − 1. Find fg(x) and gf(x), then compare them at x = 2.
Answer
fg(x) = f(x² − 1) = x² − 1 + 5 = x² + 4.
gf(x) = g(x + 5) = (x + 5)² − 1 = x² + 10x + 24.
At x = 2: fg(2) = 8 and gf(2) = 4 + 20 + 24 = 48. The order matters, so the two are different.
Questions 5 to 8: inverses and unknowns
Question 5. f(x) = 4x − 7. Find f⁻¹(x) and f⁻¹(5).
Answer
y = 4x − 7, so x = (y + 7) ÷ 4. Hence f⁻¹(x) = (x + 7) ÷ 4.
f⁻¹(5) = 12 ÷ 4 = 3. Check: f(3) = 12 − 7 = 5.
Question 6. f(x) = (x + 1) ÷ (x − 3), x ≠ 3. Find f⁻¹(x) and state the value that x cannot take in f⁻¹.
Answer
y(x − 3) = x + 1, so yx − 3y = x + 1 and x(y − 1) = 3y + 1. Hence x = (3y + 1) ÷ (y − 1).
f⁻¹(x) = (3x + 1) ÷ (x − 1), x ≠ 1.
Check: f(5) = 6 ÷ 2 = 3 and f⁻¹(3) = 10 ÷ 2 = 5.
Question 7. f(x) = (x − 3)² with domain x ≥ 3. Find f⁻¹(x) and f⁻¹(16).
Answer
Because x ≥ 3, x − 3 is not negative, so x − 3 = √y and x = 3 + √y.
f⁻¹(x) = 3 + √x, x ≥ 0. Then f⁻¹(16) = 3 + 4 = 7. Check: f(7) = 4² = 16.
Question 8. f(x) = 3x − 2 and fg(x) = 6x + 4. Find g(x), then solve f(x) = g(x).
Answer
f(g(x)) = 3g(x) − 2 = 6x + 4, so 3g(x) = 6x + 6 and g(x) = 2x + 2. Check: 3(2x + 2) − 2 = 6x + 4.
Now solve 3x − 2 = 2x + 2, which gives x = 4. Check: f(4) = 10 and g(4) = 10.
If you got these wrong
Wrong on questions 1 to 3? Revisit reading function notation, domain and range. Wrong on question 4? Go back to composite functions in the right order.
Wrong on questions 5 to 7? Read inverse functions with domain restrictions. Wrong on question 8? Use solving equations with an unknown function.
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