A7: Functions
Work with functions and their inverses; composite functions; f(x), fg(x), f⁻¹(x) notation
Work with functions and their inverses; composite functions; f(x), fg(x), f⁻¹(x) notation
Given f(x) = 3x - 2, find:
a) f(4)
b) f(-2)
c) f(a)
Solution:
a) f(4) = 3(4) - 2 = 12 - 2 = 10
b) f(-2) = 3(-2) - 2 = -6 - 2 = -8
c) f(a) = 3a - 2
Given g(x) = x² + 1, find:
a) g(3)
b) g(-3)
c) g(2x)
Solution:
a) g(3) = 3² + 1 = 10
b) g(-3) = (-3)² + 1 = 10
c) g(2x) = (2x)² + 1 = 4x² + 1
Find f⁻¹(x) when f(x) = 3x + 5
Solution:
Step 1: Let y = 3x + 5
Step 2: Make x the subject:
y - 5 = 3x
x = y - 5⁄3
Step 3: Replace y with x:
f⁻¹(x) = x - 5⁄3
Find f⁻¹(x) when f(x) = x + 3⁄2
Solution:
y = x + 3⁄2
2y = x + 3
x = 2y - 3
f⁻¹(x) = 2x - 3
Given f(x) = 2x + 1 and g(x) = x², find:
a) fg(3)
b) gf(3)
Solution:
a) fg(3) = f(g(3)) = f(3²) = f(9) = 2(9) + 1 = 19
b) gf(3) = g(f(3)) = g(2(3) + 1) = g(7) = 7² = 49
Given f(x) = x + 4 and g(x) = 2x - 1, find fg(x)
Solution:
fg(x) = f(g(x))
= f(2x - 1)
= (2x - 1) + 4
= 2x + 3
Given f(x) = 3x and g(x) = x - 2, find gf(x)
Solution:
gf(x) = g(f(x))
= g(3x)
= 3x - 2
Find f⁻¹(x) when f(x) = x² + 5, x ≥ 0
Solution:
y = x² + 5
y - 5 = x²
x = √(y - 5)
f⁻¹(x) = √(x - 5), x ≥ 5
Function machines: f(x) is the rule. Composite fg(x) means 'g FIRST, then f' — work from the inside out.
Given f(x) = 2x + 1, find f(−3).
f(−3) = 2(−3) + 1 = −6 + 1 = −5
B1 for −5. Substitution with brackets — the (−3) gets multiplied by 2, not the 3.
Find the inverse of f(x) = 3x + 6.
Rearrange: y = 3x + 6 → x = (y − 6)/3
f⁻¹(x) = (x − 6)/3
M1 for swapping/rearranging, A1 for (x − 6)/3. Check with a value: f(2) = 12 and f⁻¹(12) = 2 ✓
Exam wording to expect: "Find f(…)", "Find the inverse function", "Find fg(…)".
Q1: Given f(x) = 4x - 3, find f(5) and f(-2).
Q2: Given g(x) = x² - 1, find g(3) and g(x + 1).
Q3: Find the inverse of f(x) = 2x - 7.
Q4: Find the inverse of g(x) = x⁄4 + 1.
Q5: Given f(x) = x + 3 and g(x) = 2x, find fg(x) and gf(x).
Q6: Given f(x) = x² and g(x) = x + 1, find fg(2) and gf(2).
Q7: Given f(x) = 3x − 1 and g(x) = x², find fg(3)
Q8: Find the inverse of h(x) = (x + 2)/5
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. f(5) = 20 − 3 = 17; f(−2) = −8 − 3 = −11
Q2. g(3) = 9 − 1 = 8; g(x + 1) = (x + 1)² − 1 = x² + 2x + 1 − 1 = x² + 2x
Q3. y = 2x − 7 → x = (y + 7)/2 → f⁻¹(x) = (x + 7)/2
Q4. y = x/4 + 1 → x = 4(y − 1) → g⁻¹(x) = 4x − 4
Q5. fg(x) = f(2x) = 2(2x) + 3 = 4x + 3; gf(x) = g(x + 3) = 2(x + 3) = 2x + 6
Q6. fg(2) = f(3) = 9; gf(2) = g(4) = 5 — order matters!
Q7. g(3) = 9, then f(9) = 27 − 1 = 26
Q8. y = (x + 2)/5 → x = 5y − 2 → h⁻¹(x) = 5x − 2
Functions: 2–4 marks. Inverses appear on Paper 1; composites on Paper 2/3. The "which order?" trap in composites is the discriminator.
Links to: substitution (A2), rearranging (A5), solving equations (A17).
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
Given f(x) = 2x + 1 and g(x) = x². (a) Find ff(3). (b) Find f⁻¹(x). (c) Solve fg(x) = 33.
Solution:
(a) f(3) = 7, then f(7) = 15
(b) y = 2x + 1 → x = y - 1⁄2, so f⁻¹(x) = x - 1⁄2
(c) fg(x) = f(x²) = 2x² + 1 = 33 → 2x² = 32 → x² = 16 → x = ±4
1. Wrong: fg(x) means f × g(x) Correct: fg(x) = f(g(x)) — apply g first, then f to the result
2. Wrong: f⁻¹(x) means 1⁄f(x) Correct: f⁻¹(x) is the inverse function, not the reciprocal
3. Wrong: fg(x) = gf(x) always Correct: Order matters! fg(x) ≠ gf(x) in general
6 marks: f(x) = 3x - 5 and g(x) = x + 2⁄4. (a) Find f⁻¹(x). (b) Find gf(3). (c) Show that fg(x) ≠ gf(x) by finding both expressions.
(a) y = 3x - 5 → y + 5 = 3x → x = y + 5⁄3, so f⁻¹(x) = x + 5⁄3
(b) f(3) = 3(3) - 5 = 4. g(4) = 4 + 2⁄4 = 6⁄4 = 1.5
(c) fg(x) = f(x + 2⁄4) = 3(x + 2⁄4) - 5 = 3x + 6⁄4 - 5 = 3x - 14⁄4
gf(x) = g(3x - 5) = 3x - 5 + 2⁄4 = 3x - 3⁄4
These are different: 3x - 14⁄4 ≠ 3x - 3⁄4
Mark scheme: (a) 2 marks. (b) 1 mark for f(3), 1 mark for final answer. (c) 1 mark for fg(x), 1 mark for gf(x).
f(x) = 2x + 3 converts a temperature from °C to an adjusted scale. g(x) = x - 3⁄2 is its inverse.
(a) What does f(0) represent?
(b) If the output of f is 15, what was the input?
(c) Explain why g undoes f.
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