Printable · GCSE Higher · ages 14-16
Functions, inverse and composite functions worksheet — GCSE Higher
Fifteen questions on "functions, inverse and composite functions" — DfE statement A7. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Answer key: Functions, inverse and composite functions worksheet — GCSE Higher
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- (b) 143 — fg(4) means f(g(4)): work out g(4) first, then substitute the result into f. g(4) = 3 × 4 = 12, then f(12) = 12² − 1 = 144 − 1 = 143. Working out gf(4) instead swaps the order: f(4) = 4² − 1 = 15, then g(15) = 3 × 15 = 45 — that is the wrong composition. Treating f(x) as x − 1 (forgetting to square the input) gives f(12) = 12 − 1 = 11. Applying g twice instead of applying g then f gives g(g(4)) = g(12) = 3 × 12 = 36, which mixes up which function should be applied second.
- (d) 11 — f⁻¹(x) = 5 means x = f(5), since applying f to both sides undoes the inverse. f(5) = 2 × 5 + 1 = 11. Writing 2 comes from confusing f⁻¹(x) = 5 with f(x) = 5, and solving 2x + 1 = 5 instead: 2x = 4, x = 2. Writing 9 comes from finding f⁻¹(x) with a sign error, f⁻¹(x) = (x + 1)/2 instead of (x − 1)/2, then setting this equal to 5: x + 1 = 10, x = 9. Writing 6 comes from finding f⁻¹(x) without dividing by 2 at all, f⁻¹(x) = x − 1, then setting this equal to 5: x = 6.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (a) y = −2x + 3 — Method: find the step in the outputs for each step of 1 in the input — falling outputs mean a negative multiplier — then read off the output when the input is 0, because that is the number added on. Working: the outputs 5, 3, 1, −1 fall by 2 each time x rises by 1, so x is multiplied by −2; the output at x = 0 is 3, so 3 is added. Answer: y = −2x + 3, checked at x = 2 by −2 × 2 + 3 = −1. The distractors: y = 2x + 3 comes from taking the size of the step, 2, as the multiplier and ignoring the fact that the outputs are falling; y = −2x − 3 comes from using the correct multiplier but writing the number added on as −3 instead of the output 3 listed at x = 0; y = −x + 4 comes from taking the multiplier as −1, its size read from the step of 1 in the inputs instead of the step of 2 in the outputs and its sign from the fact that the outputs fall, and then fitting the number added on to the pair x = −1, y = 5.
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
- (b) 2x + 3 — fg(x) means f(g(x)): apply g first, then apply f to the result. g(x) = 2x, so f(g(x)) = f(2x) = 2x + 3. Writing 2x + 6 comes from working out gf(x) instead — g(f(x)) = g(x + 3) = 2(x + 3) = 2x + 6 — which applies the functions in the wrong order. Writing 3x + 3 comes from adding f(x) and g(x) together, (x + 3) + 2x = 3x + 3, instead of composing them. Writing 2x² + 6x comes from multiplying f(x) and g(x) together, (x + 3)(2x) = 2x² + 6x, instead of substituting one into the other.
- (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
- (a) 2x − 1 — Swap x and y: x = (y + 1)/2. Multiply both sides by 2: 2x = y + 1. Subtract 1 from both sides: y = 2x − 1, so f⁻¹(x) = 2x − 1. Writing 2x + 1 comes from not flipping the sign on the 1 when it is moved across the equals sign. Writing (x − 1)/2 comes from reversing the sign of the 1 but leaving the ÷2 from the original rule in place, instead of turning it into ×2. Writing x/2 − 1 comes from dividing only the x by 2 and treating the 1 as already outside the fraction.
- (c) Correct: 3y = 4 − x gives f⁻¹(x) = (4 − x)/3 — Swap x and y: x = 4 − 3y. Add 3y to both sides and subtract x from both sides: 3y = 4 − x. Divide by 3: y = (4 − x)/3, which is exactly what Ben wrote — his rearrangement is correct. Check with a value: f(1) = 4 − 3 = 1, and Ben's formula gives (4 − 1)/3 = 1, which matches. 'Correct, but only because f is its own inverse' gives the right verdict for a false reason — f(f(x)) = 4 − 3(4 − 3x) = 9x − 8, which is not x, so f is not self-inverse; Ben's rearrangement is correct for the ordinary algebraic reason above, not because of any special property of f. 'Wrong: sign kept, giving (−4 − x)/3' comes from not carrying the swap through consistently — testing x = 1 gives (−4 − 1)/3 = −5/3, which does not equal 1, so it is wrong. 'Wrong: correct inverse is (x − 4)/3' comes from writing 3y = x − 4 instead of 3y = 4 − x, a sign slip when isolating y — testing x = 1 gives (1 − 4)/3 = −1, which again does not equal 1.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (c) −0.5 — fg(x) = f(g(x)) = f(x²) = x² + 2. gf(x) = g(f(x)) = g(x + 2) = (x + 2)² = x² + 4x + 4. Setting fg(x) = gf(x): x² + 2 = x² + 4x + 4. Subtract x² from both sides: 2 = 4x + 4. Subtract 4 from both sides: −2 = 4x, so x = −0.5. Writing 1.5 comes from adding the 4 instead of subtracting it: 4x = 2 + 4 = 6, giving x = 1.5. Writing 'no solution' comes from expanding (x + 2)² as x² + 4 using (a + b)² = a² + b², losing the middle term — the equation then reads x² + 2 = x² + 4, which has no solution, but the expansion itself is wrong. Writing 0 comes from treating gf(x) as g(x) + f(x) instead of g(f(x)): x² + (x + 2) = x² + 2 gives x = 0, but that adds the two functions rather than composing them.
- (c) 4x − 3 — ff(x) means f(f(x)): substitute f(x) into f in place of x. f(f(x)) = 2 × f(x) − 1 = 2 × (2x − 1) − 1. Expanding the bracket: 2 × (2x − 1) = 4x − 2. Combining the constant terms: −2 − 1 = −3, so f(f(x)) = 4x − 3. Writing 4x − 2 comes from expanding 2(2x − 1) correctly to get 4x − 2, then forgetting to subtract the outer 1 at all. Writing 4x² − 4x + 1 comes from reading ff(x) as f(x) multiplied by itself, (2x − 1)(2x − 1) = 4x² − 4x + 1, instead of substituting f(x) into f. Writing 4x − 1 comes from doubling the coefficient of x in the original rule directly, without actually substituting f(x) into f at all.
- (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
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