Every functions question from 8 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.13. 8 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/42/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.
0580 Functions past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q17
f(x) = 20 − 3x, g(x) = 64⁄2ˣ, h(x) = 7 + 2x.
(a) Find f(4). (b) Find h⁻¹(x). (c) Find gg(5). (d) Solve the equation g⁻¹(x) = 8.
Show solution
- Substitute x = 4: f(4) = 20 − 3 × 4 = 20 − 12 = 8
- Write y = 7 + 2x and make x the subject.
- Subtract 7: y − 7 = 2x. Divide by 2: x = y − 7⁄2.
- Swap y for x: h⁻¹(x) = x − 7⁄2
- Work from the inside out. g(5) = 64⁄2⁵ = 64⁄32 = 2.
- Now put 2 into g: g(2) = 64⁄2² = 64⁄4 = 16
- g⁻¹(x) = 8 means “the input to g that gives x is 8”, so x = g(8).
- g(8) = 64⁄2⁸ = 64⁄256 = 1⁄4
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q18
(a) f(x) = 3x + 1 and g(x) = 5 − 3x. (i) Find x when f(x) = g(x). (ii) Find g⁻¹(x). (iii) Find gg(x). Give your answer in its simplest form.
(b) h(x) = 2²ˣ. Find x when h⁻¹(x) = 2.
Show solution
- Set the two functions equal: 3x + 1 = 5 − 3x.
- Add 3x to both sides: 6x + 1 = 5. Subtract 1: 6x = 4.
- x = 4⁄6 = 2⁄3
- Write y = 5 − 3x and make x the subject.
- Add 3x and subtract y: 3x = 5 − y, so x = 5 − y⁄3.
- Swap y for x: g⁻¹(x) = 5 − x⁄3
- gg(x) means put g(x) into g: gg(x) = 5 − 3(5 − 3x).
- Expand: 5 − 15 + 9x.
- 9x − 10
- h⁻¹(x) = 2 means that the input 2 into h gives x. So x = h(2).
- h(2) = 22 × 2 = 2⁴ = 16
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q15
(a) f(x) = 10 − 2x and g(x) = 2x² + 3. (i) The domain of f(x) is {−2, 0, 3}. Find the range of f(x). (ii) Find f⁻¹(x). (iii) Find gf(x) in the form ax² + bx + c.
(b) h(x) = 102/x, x ≠ 0. Find h⁻¹(10 000).
Show solution
- Substitute each value in the domain into f(x) = 10 − 2x.
- f(−2) = 10 + 4 = 14. f(0) = 10. f(3) = 10 − 6 = 4.
- Range = {14, 10, 4}
- Write y = 10 − 2x and make x the subject.
- Add 2x and subtract y: 2x = 10 − y, so x = 10 − y⁄2.
- Swap y for x: f⁻¹(x) = 10 − x⁄2
- gf(x) means put f(x) into g: gf(x) = 2(10 − 2x)² + 3.
- Expand the square: (10 − 2x)² = 100 − 40x + 4x².
- Multiply by 2 and add 3: 200 − 80x + 8x² + 3.
- 8x² − 80x + 203
- h⁻¹(10 000) is the x-value that gives h(x) = 10 000.
- Write 10 000 as a power of 10: 10 000 = 10⁴. So 102/x = 10⁴.
- The powers must be equal: 2⁄x = 4, so 2 = 4x and x = 1⁄2
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q18
f(x) = 2x + 5. f(x)f(x) − ff(x) = ax² + bx + c. Find the value of a, the value of b and the value of c.
Show solution
- f(x)f(x) means f(x) multiplied by itself: (2x + 5)(2x + 5).
- Expand: 4x² + 10x + 10x + 25 = 4x² + 20x + 25.
- ff(x) means put f(x) into f: f(2x + 5) = 2(2x + 5) + 5 = 4x + 10 + 5 = 4x + 15.
- Subtract: (4x² + 20x + 25) − (4x + 15) = 4x² + 16x + 10.
- a = 4, b = 16, c = 10
Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q20
f(x) = 21⁄2x − 1, x ≠ 1⁄2, and g(x) = 3x + 4.
(a) Find (i) g(2) (ii) gf(−1) (iii) f⁻¹(x).
(b) Solve f(x) = g(x).
Show solution
- Substitute x = 2 into g(x) = 3x + 4: 3 × 2 + 4 = 10
- gf(−1) means do f first, then g.
- f(−1) = 21⁄2(−1) − 1 = 21⁄−3 = −7.
- g(−7) = 3 × (−7) + 4 = −21 + 4 = −17
- Write y = 21⁄2x − 1 and make x the subject.
- Multiply by (2x − 1): y(2x − 1) = 21, so 2xy − y = 21.
- Add y: 2xy = 21 + y. Divide by 2y: x = 21 + y⁄2y.
- Swap the letters: f⁻¹(x) = 21 + x⁄2x
- Set f(x) = g(x): 21⁄2x − 1 = 3x + 4.
- Multiply both sides by (2x − 1): 21 = (3x + 4)(2x − 1) = 6x² − 3x + 8x − 4 = 6x² + 5x − 4.
- Rearrange: 6x² + 5x − 25 = 0.
- Factorise: (3x − 5)(2x + 5) = 0.
- x = 5⁄3 or x = −5⁄2
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q28
f(x) = 7x − 4. Find the value of x when (a) f(x) = 1 (b) f⁻¹(x) = 1.
Show solution
- Any number to the power 0 equals 1, so 1 = 7⁰.
- 7x − 4 = 7⁰ means x − 4 = 0.
- x = 4
- f⁻¹(x) = 1 means “f sends 1 to x”, so x = f(1).
- f(1) = 71 − 4 = 7⁻³ = 1⁄7³
- x = 1⁄343
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q19
f(x) = 5ˣ, g(x) = 3x − 2 and h(x) = x² + 1.
(a) Find f(5). (b) Find g(8x). (c) Find g⁻¹(x). (d) Find the positive solution of gh(x) = 364. (e) Find ff⁻¹(12).
Show solution
- f(5) = 5⁵ = 5 × 5 × 5 × 5 × 5 = 3125
- Replace x with 8x in g(x) = 3x − 2: g(8x) = 3 × 8x − 2 = 24x − 2
- Write y = 3x − 2 and make x the subject.
- Add 2: y + 2 = 3x. Divide by 3: x = y + 2⁄3.
- Swap y for x: g⁻¹(x) = x + 2⁄3
- gh(x) means put h(x) into g: gh(x) = 3(x² + 1) − 2 = 3x² + 3 − 2 = 3x² + 1.
- Set it equal to 364: 3x² + 1 = 364, so 3x² = 363 and x² = 121.
- x = ±11. The positive solution is x = 11
- f⁻¹ undoes f, so applying f after f⁻¹ gives back the starting number.
- ff⁻¹(12) = 12
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q14
f(x) = 3x − 4 and g(x) = 4x + 1
(a) Find f(−2).
(b) Find f−1(x).
(c) fg(x) = ax + b. Find the value of a, and the value of b.
(d) Simplify 2⁄f(x) − 5⁄g(x). Give your answer as a single fraction in terms of x.
Show solution
- Substitute x = −2: f(−2) = 3(−2) − 4 = −6 − 4 = −10
- Write y = 3x − 4 and make x the subject: x = y + 4⁄3.
- Swap the letters to write the inverse in terms of x.
- f−1(x) = x + 4⁄3
- fg(x) means do g first, then f: fg(x) = f(4x + 1).
- f(4x + 1) = 3(4x + 1) − 4 = 12x + 3 − 4 = 12x − 1.
- a = 12, b = −1
- Write the expression: 2⁄3x − 4 − 5⁄4x + 1.
- Use the common denominator (3x − 4)(4x + 1): 2(4x + 1) − 5(3x − 4)⁄(3x − 4)(4x + 1)
- Expand the top: 8x + 2 − 15x + 20 = 22 − 7x.
- 22 − 7x⁄(3x − 4)(4x + 1)
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