Advanced Functions – Grade 10 IGCSE Maths

Welcome to Advanced Functions!

Functions are one of the most powerful tools in mathematics. In this topic you will master composite functions (combining two functions), inverse functions (reversing a function), and domain & range (understanding valid inputs and outputs). These ideas underpin all of A-Level mathematics.

f∘g(x) = fg(x) = f(g(x)) | f⁻¹ reverses f | ff⁻¹(x) = x

Composite Functions

fg(x) = f(g(x)) — apply g first, then f

Inverse Functions

Swap x↔y and rearrange for f⁻¹(x)

Domain & Range

Valid inputs and possible outputs

Worked Examples

6 step-by-step examples

Visualiser

Function machine — plot fg and gf

Exercises

5 sets × 8 questions + Practice + Challenge

1. Composite Functions

A composite function is one function applied to the output of another. The notation fg(x) means "apply g first, then apply f to the result."

fg(x) = f(g(x)) — evaluate g(x) first, then substitute into f

Example: f(x) = 2x + 1, g(x) = x²
fg(x) = f(g(x)) = f(x²) = 2(x²) + 1 = 2x² + 1
gf(x) = g(f(x)) = g(2x+1) = (2x+1)² = 4x² + 4x + 1

Notice: fg(x) ≠ gf(x) — order matters!

Evaluating fg(3):
Step 1: g(3) = 3² = 9
Step 2: f(9) = 2(9) + 1 = 19

Alternatively: fg(3) = 2(3²) + 1 = 2(9) + 1 = 19 ✓

Memory trick: fg(x) — read right to left. g is closer to x so g acts first. Think of it like getting dressed: socks (g) go on before shoes (f).

Three-Function Composition

For fgh(x), apply h first, then g, then f — always work right to left.

f(x) = x+1, g(x) = 2x, h(x) = x²:
fgh(x) = f(g(h(x))) = f(g(x²)) = f(2x²) = 2x² + 1
fgh(3) = h(3)=9, g(9)=18, f(18)=19 → 19

Solving fg(x) = k

Solve fg(x) = 19 where f(x)=2x+1, g(x)=x²:
fg(x) = 2x² + 1 = 19
2x² = 18 → x² = 9 → x = ±3

2. Inverse Functions

The inverse function f⁻¹(x) reverses the effect of f(x). If f maps x to y, then f⁻¹ maps y back to x.

Method: write y = f(x) → swap x and y → rearrange to make y the subject → that's f⁻¹(x)

Find f⁻¹(x) for f(x) = 3x − 2:
Step 1: y = 3x − 2
Step 2: swap: x = 3y − 2
Step 3: rearrange: x + 2 = 3y → y = (x+2)/3
So: f⁻¹(x) = (x+2)/3

Verify: f(f⁻¹(x)) = 3·(x+2)/3 − 2 = x+2−2 = x ✓

Find f⁻¹(x) for f(x) = (2x+1)/(x−3):
y = (2x+1)/(x−3)
Swap: x = (2y+1)/(y−3)
Multiply: x(y−3) = 2y+1
xy − 3x = 2y + 1
xy − 2y = 3x + 1
y(x−2) = 3x + 1
f⁻¹(x) = (3x+1)/(x−2)

Key fact: The graph of f⁻¹(x) is the reflection of the graph of f(x) in the line y = x. The domain of f⁻¹ equals the range of f, and the range of f⁻¹ equals the domain of f.

One-to-one requirement: An inverse only exists if the function is one-to-one (each output comes from exactly one input). f(x) = x² is NOT one-to-one over all reals, so its inverse only exists when the domain is restricted to x ≥ 0.

3. Domain, Range and Restrictions

The domain is the set of all valid input values (x-values). The range is the set of all possible output values (y-values).

f(x) = 1/(x−2):
Domain: x ≠ 2 (cannot divide by zero)
Range: f(x) ≠ 0 (the fraction can never equal zero)

f(x) = √(x+3):
Domain: x+3 ≥ 0 → x ≥ −3
Range: f(x) ≥ 0 (square root is never negative)

Domain of fg(x): first need g(x) to be in the domain of f

Example: f(x) = √x, g(x) = x−4 — find domain of fg(x):
fg(x) = f(g(x)) = √(x−4)
Need: x−4 ≥ 0 → domain of fg is x ≥ 4

Piecewise function example:
f(x) = 2x + 1 for x ≥ 0; f(x) = x² for x < 0
f(3) = 2(3)+1 = 7; f(−2) = (−2)² = 4
Domain: all real numbers; Range: f(x) ≥ 0

Watch out for: division by zero (exclude that value), square roots of negatives (need ≥ 0 inside), logarithms (need > 0 inside).

Example 1 — Find fg(x) and gf(x)

Given: f(x) = 3x + 2, g(x) = x² − 1

fg(x) = f(g(x)) = f(x²−1) = 3(x²−1) + 2 = 3x² − 3 + 2 = 3x² − 1

gf(x) = g(f(x)) = g(3x+2) = (3x+2)² − 1 = 9x² + 12x + 4 − 1 = 9x² + 12x + 3

Note: fg ≠ gf in general — always check the order carefully.

Example 2 — Evaluate fg(a) for specific value

Given: f(x) = 2x − 5, g(x) = x + 3. Find fg(4).

Step 1: g(4) = 4 + 3 = 7

Step 2: f(7) = 2(7) − 5 = 14 − 5 = 9

Check via formula: fg(x) = 2(x+3)−5 = 2x+1; fg(4) = 9 ✓

Example 3 — Find inverse of a linear function

Given: f(x) = 5x − 3. Find f⁻¹(x).

y = 5x − 3 → swap: x = 5y − 3

x + 3 = 5y → y = (x+3)/5

f⁻¹(x) = (x+3)/5

Example 4 — Inverse of a fractional function

Given: f(x) = (x+1)/(x−2). Find f⁻¹(x).

x = (y+1)/(y−2) → x(y−2) = y+1 → xy−2x = y+1

xy − y = 2x + 1 → y(x−1) = 2x+1

f⁻¹(x) = (2x+1)/(x−1)

Example 5 — Domain and Range

f(x) = √(2x−6): need 2x−6 ≥ 0 → x ≥ 3

Domain: x ≥ 3. Range: f(x) ≥ 0.

g(x) = 3/(x+5): need x+5 ≠ 0 → x ≠ −5. Range: g(x) ≠ 0.

Example 6 — Solve fg(x) = k

f(x) = x + 3, g(x) = x². Solve fg(x) = 12.

fg(x) = x² + 3 = 12 → x² = 9 → x = ±3

Common Mistakes

Mistake 1: fg(x) = f(x) × g(x)
Wrong! fg means composition — apply g first, then f. It is NOT multiplication. f(x)·g(x) is written f(x)g(x) or [f(x)][g(x)], never fg(x).
Mistake 2: Reversing the order — thinking fg means f first
fg(x) = f(g(x)) — g is applied FIRST. The function nearest to x is always applied first. Many students apply f first and get the wrong answer.
Mistake 3: Inverse rearrangement error — not swapping x and y
You MUST swap x and y before rearranging. A common error is to just rearrange y=f(x) for x without swapping, which gives the wrong expression.
Mistake 4: Domain errors — forgetting to exclude values
For 1/(x−3), the domain is x ≠ 3, not x > 3. For √(x−5), you need x−5 ≥ 0, so domain is x ≥ 5, not x > 5 (equality is included).
Mistake 5: Range of f⁻¹ = domain of f (not range of f)
The domain and range swap when you take an inverse. Domain of f⁻¹ = Range of f. Range of f⁻¹ = Domain of f. Don't mix these up.

Key Formulas & Notation

Concept	Notation / Formula	Meaning
Composite function	fg(x) or f∘g(x)	f(g(x)) — apply g first, then f
Evaluate composite	fg(a): g(a) first, then f(result)	Work step by step
Inverse function	f⁻¹(x)	Reverses f: write y=f(x), swap x↔y, rearrange for y
Verify inverse	ff⁻¹(x) = x and f⁻¹f(x) = x	Both compositions give x
Domain restriction	f(x)=1/(x−a): x ≠ a	Exclude value making denominator zero
Square root domain	f(x)=√(g(x)): g(x) ≥ 0	Expression under root must be non-negative
Domain of fg	Need g(x) in domain of f	g(x) must produce values f can accept
Inverse graph	f⁻¹ is reflection of f in y = x	Coordinates swap: (a,b) → (b,a)

Function Machine Visualiser

Enter simple polynomial expressions for f and g (use x as variable, e.g. "2*x+1" or "x*x"). The visualiser will plot fg(x) in pink and gf(x) in blue, and compute f⁻¹ for linear f.

f(x) = g(x) =

Enter expressions and click Plot.

Exercise 1 — Evaluating Composite Functions

f(x) = 2x + 3, g(x) = x². Give exact numeric answers.

1. Find fg(2). [fg(x)=2x²+3; fg(2)=2(4)+3]

2. Find gf(2). [gf(x)=(2x+3)²; gf(2)=(7)²]

3. Find fg(3).

4. Find gf(1). [gf(1)=(2+3)²=25]

5. Find fg(0). [fg(0)=2(0)+3=3]

6. Find gf(0). [gf(0)=(3)²=9]

7. Find fg(−1). [fg(−1)=2(1)+3=5]

8. Find gf(−1). [gf(−1)=(−2+3)²=(1)²=1]

Exercise 2 — Inverse Functions

Find f⁻¹(a) for the given value. Show your working on paper.

1. f(x) = x + 5. Find f⁻¹(8). [f⁻¹(x)=x−5]

2. f(x) = 2x − 1. Find f⁻¹(7). [f⁻¹(x)=(x+1)/2]

3. f(x) = 3x + 6. Find f⁻¹(0). [f⁻¹(x)=(x−6)/3]

4. f(x) = 4x − 8. Find f⁻¹(4). [f⁻¹(x)=(x+8)/4]

5. f(x) = (x+3)/2. Find f⁻¹(5). [f⁻¹(x)=2x−3]

6. f(x) = x/3 + 2. Find f⁻¹(5). [f⁻¹(x)=3(x−2)=3x−6]

7. f(x) = 5x + 10. Find f⁻¹(−5). [f⁻¹(x)=(x−10)/5]

8. f(x) = 2x + 4. Find f⁻¹(10). [f⁻¹(x)=(x−4)/2]

Exercise 3 — Domain and Range

Enter the boundary or excluded value as a number.

1. f(x)=1/(x−4). Enter the excluded domain value.

2. f(x)=√(x−7). Enter the minimum domain value.

3. f(x)=1/(x+3). Enter the excluded domain value.

4. f(x)=√(x+9). Enter the minimum domain value.

5. f(x)=√(3x−12). Enter the minimum domain value.

6. f(x)=1/(2x−10). Enter the excluded domain value.

7. fg(x) where f(x)=√x and g(x)=x−6. Enter minimum domain value for fg.

8. fg(x) where f(x)=1/x and g(x)=x−5. Enter excluded domain value for fg.

Exercise 4 — Solving fg(x) = k

Find the positive solution for x (or the single solution given).

1. f(x)=x+1, g(x)=x². Solve fg(x)=5. Give positive x. [x²+1=5 → x²=4]

2. f(x)=2x, g(x)=x+3. Solve fg(x)=10. [2(x+3)=10 → x+3=5]

3. f(x)=x−4, g(x)=x². Solve fg(x)=0. Give positive x. [x²−4=0]

4. f(x)=3x+1, g(x)=x−2. Solve fg(x)=7. [3(x−2)+1=7 → 3x−5=7]

5. f(x)=x², g(x)=2x+1. Solve gf(x)=9. Give positive x. [(2x+1)=9 → nope: gf(x)=(2x+1)² wait — gf means f first... gf(x)=g(f(x))=g(x²)=2x²+1=9 → x²=4]

6. f(x)=x+2, g(x)=x². Solve fg(x)=11. Give positive x. [x²+2=11 → x²=9]

7. f(x)=4x, g(x)=x−1. Solve fg(x)=12. [4(x−1)=12 → x−1=3]

8. f(x)=x+5, g(x)=x². Solve fg(x)=14. Give positive x. [x²+5=14 → x²=9]

Exercise 5 — Combined and Three-Function Composition

f(x)=x+1, g(x)=2x, h(x)=x². Give exact numeric answers.

1. Find fgh(2). [h(2)=4, g(4)=8, f(8)=9]

2. Find fgh(3). [h(3)=9, g(9)=18, f(18)=19]

3. Find ghf(2). [f(2)=3, h(3)=9, g(9)=18]

4. Find hgf(1). [f(1)=2, g(2)=4, h(4)=16]

5. Find fhg(2). [g(2)=4, h(4)=16, f(16)=17]

6. Find ggf(1). g∘g∘f: f(1)=2, g(2)=4, g(4)=8]

7. Find ffg(3). [g(3)=6, f(6)=7, f(7)=8]

8. Find hfg(2). [g(2)=4, f(4)=5, h(5)=25]

Practice — 25 Questions

Mixed questions. f(x)=3x+1, g(x)=x²−2, h(x)=2x unless stated.

1. fg(2): g(2)=2, f(2)=3(2)+1 → fg(2)=3(x²−2)+1 at x=2 → 3(2)+1

2. fg(3): g(3)=7, f(7)=22

3. gf(2): f(2)=7, g(7)=47

4. gf(1): f(1)=4, g(4)=14

5. hf(3): f(3)=10, h(10)=20

6. fh(4): h(4)=8, f(8)=25

7. f⁻¹(x)=(x−1)/3. Find f⁻¹(10).

8. f⁻¹(4).

9. f⁻¹(−2).

10. p(x)=2x−8. Find p⁻¹(0). [p⁻¹(x)=(x+8)/2]

11. p⁻¹(6). [(6+8)/2=7]

12. q(x)=x/4+3. Find q⁻¹(5). [q⁻¹(x)=4(x−3)]

13. q⁻¹(7).

14. f(x)=1/(x−5). Excluded domain value?

15. f(x)=√(x−9). Minimum domain value?

16. f(x)=1/(x+7). Excluded domain value?

17. fg(x)=x²+1 with f(x)=x+1. Solve fg(x)=10. Give positive x. [x²=9]

18. fg(x)=2x²+1. Solve fg(x)=19. Give positive x. [2x²=18 → x²=9]

19. gf(x)=(3x+1)²−2. Find gf(0). [(1)²−2=−1]

20. gf(−1). gf(−1)=(−2)²−2=2

21. fgh(2): h(2)=4, g(4)=14, f(14)=43

22. fgh(1): h(1)=1, g(1)=−1, f(−1)=−2

23. s(x)=4x−12. Find s⁻¹(0). [s⁻¹(x)=(x+12)/4]

24. s⁻¹(8). [(8+12)/4=5]

25. f(x)=√(2x−4). Minimum domain value? [2x≥4 → x≥2]

Challenge — 12 Questions

Harder problems involving inverse of fractional functions and combined domain restrictions.

1. f(x)=(2x+1)/(x−3). Find f⁻¹(x). Evaluate f⁻¹(5). [y=(3x+1)/(x−2); f⁻¹(5)=(16)/(3)≈5.33]

2. f(x)=(x+4)/(x−1). Find f⁻¹(3). [f⁻¹(x)=(x+4)/(x−1); swap: x=(y+4)/(y−1) → y=(x+4)/(x−1); f⁻¹(3)=7/2=3.5]

3. f(x)=3x−7, g(x)=x²+1. Solve fg(x)=20. Give positive x. [3(x²+1)−7=20 → 3x²=24 → x²=8 → x=2√2≈2.83]

4. f(x)=(5x−1)/(2). Find f⁻¹(7). [f⁻¹(x)=(2x+1)/5; (14+1)/5=3]

5. f(x)=2x+3, g(x)=√x. Domain of gf(x)? Enter minimum value. [gf(x)=√(2x+3); 2x+3≥0 → x≥−1.5]

6. f(x)=x+2, g(x)=x², h(x)=3x. Find fgh(1). [h(1)=3, g(3)=9, f(9)=11]

7. f(x)=(x−1)/(x+2). Find f⁻¹(0). [x=(y−1)/(y+2) → x(y+2)=y−1 → y(x−1)=−2x−1 → y=(−2x−1)/(x−1); f⁻¹(0)=(0−1)/(0−1)=1]

8. p(x)=x²−5, q(x)=2x+1. Solve pq(x)=−1. Give positive x. [p(2x+1)=(2x+1)²−5=−1 → (2x+1)²=4 → 2x+1=2 → x=0.5]

9. f(x)=1/(2x+6). Excluded domain value? [2x+6=0 → x=−3]

10. f(x)=√(4−x). Maximum domain value? [4−x≥0 → x≤4; boundary=4]

11. f(x)=2x+1, g(x)=3x−2. Find fg⁻¹(4). [g⁻¹(x)=(x+2)/3; g⁻¹(4)=2; f(2)=5]

12. f(x)=x³. Find f⁻¹(27). [f⁻¹(x)=∛x; f⁻¹(27)=3]

Exam Style — 5 Questions

Cambridge IGCSE style questions. Show all working on paper. Enter numeric answers.

[3 marks] f(x) = 4x − 3 and g(x) = x² + 2.
(a) Find fg(x) as an expression. The coefficient of x² in fg(x) is 4.
Enter the coefficient of x².

[3 marks] f(x) = 5x + 2. Find f⁻¹(x) and evaluate f⁻¹(17). [f⁻¹(x)=(x−2)/5; f⁻¹(17)=15/5=3]

[4 marks] f(x) = (3x+1)/(x−2). Find f⁻¹(x). Evaluate f⁻¹(4). [swap: x=(3y+1)/(y−2) → x(y−2)=3y+1 → xy−2x=3y+1 → y(x−3)=2x+1 → y=(2x+1)/(x−3); f⁻¹(4)=9/1=9]

[3 marks] f(x) = 2x − 6 and g(x) = x². Solve fg(x) = 12. Give positive x. [2x²−6=12 → 2x²=18 → x²=9 → x=3]

[3 marks] f(x) = 1/(x+k) has excluded domain value x = −4. State the value of k. [x+k=0 when x=−4 → k=4]