Composite and Inverse Functions

Combining functions and finding their inverses

SPEC 2.8 • FUNCTIONS

Composite Functions

Definition

A composite function is formed when one function is applied to the result of another function.

If we have functions f and g, we can create composite functions:

fg(x)

Apply g first, then apply f to the result

fg(x) = f(g(x))

gf(x)

Apply f first, then apply g to the result

gf(x) = g(f(x))

Critical

Order matters! In general, fg(x) ≠ gf(x). The function closest to x is applied first.

Understanding the Notation

Think of composite functions as a "function machine":

Process for fg(x)

x → [g] → g(x) → [f] → f(g(x)) = fg(x)

Memory Aid

In fg(x), read right-to-left: "g of x, then f of that"

The function closest to x is applied first.

Special Notation

f²(x) means f(f(x)) — apply f twice

f³(x) means f(f(f(x))) — apply f three times

01Worked Example

Given f(x) = 2x + 3 and g(x) = x − 1, find:

(a) fg(x) (b) gf(x) (c) f²(x)

(a)

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

(b)

gf(x) = g(f(x)) = g(2x + 3) = (2x + 3) − 1 = 2x + 2

(c)

f²(x) = f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9

Note: fg(x) = 2x + 1 and gf(x) = 2x + 2 are different!

Worked Example02

Given f(x) = x² and g(x) = 3x − 2, find fg(x) and evaluate fg(2).

1

Find fg(x): fg(x) = f(g(x)) = f(3x − 2) = (3x − 2)²

2

Expand: fg(x) = (3x − 2)² = 9x² − 12x + 4

3

Evaluate fg(2): Either substitute into fg(x) or step-by-step:

•

Method 1: fg(2) = 9(4) − 12(2) + 4 = 36 − 24 + 4 = 16

•

Method 2: g(2) = 3(2) − 2 = 4, then f(4) = 4² = 16 ✓

Domain of Composite Functions

Rule

For fg(x) to be defined:

x must be in the domain of g

g(x) must be in the domain of f

Example

If f(x) = √x (domain x ≥ 0) and g(x) = x − 3

For fg(x) = √(x − 3): need x − 3 ≥ 0, so domain: x ≥ 3

Watch Out

When functions have restricted domains, carefully check that the output of the first function lies within the domain of the second.

Worked Example03

f(x) = √(x − 2), x ≥ 2 and g(x) = x². Find the domain of fg.

1

fg(x) = f(g(x)) = f(x²) = √(x² − 2)

2

For fg to be defined, we need g(x) in domain of f.

3

Domain of f is x ≥ 2, so need g(x) = x² ≥ 2

4

Solving x² ≥ 2: x ≤ −√2 or x ≥ √2

5

Domain of fg: x ≤ −√2 or x ≥ √2

Inverse Functions

Definition

The inverse function f⁻¹ "undoes" what f does. If f(a) = b, then f⁻¹(b) = a

Method to Find f⁻¹(x)

Step 1: Write y = f(x)

Step 2: Swap x and y

Step 3: Rearrange to make y the subject

Step 4: Write f⁻¹(x) = ...

Key Properties

ff⁻¹(x) = f⁻¹f(x) = x

Domain of f⁻¹ = Range of f

Range of f⁻¹ = Domain of f

Warning

f^{- 1} (x) \neq = \frac{1}{f ( x )}

The −1 means "inverse", NOT "reciprocal"!

Worked Example04

Find the inverse of f(x) = 3x − 5.

1

Write y = 3x − 5

2

Swap x and y: x = 3y − 5

3

Rearrange for y: x + 5 = 3y, so y = (x + 5)/3

4

Therefore: f⁻¹(x) = (x + 5)/3

Check: f(2) = 3(2) − 5 = 1, and f⁻¹(1) = (1 + 5)/3 = 2 ✓

Worked Example05

Find the inverse of f(x) = (x + 4)/2.

1

Write y = (x + 4)/2

2

Swap x and y: x = (y + 4)/2

3

Rearrange: 2x = y + 4, so y = 2x − 4

4

Therefore: f⁻¹(x) = 2x − 4

Verify: ff⁻¹(x) = f(2x − 4) = ((2x − 4) + 4)/2 = 2x/2 = x ✓

Domain and Range of Inverses

Key Relationships

Domain of f⁻¹ = Range of f

Range of f⁻¹ = Domain of f

Why?

If f takes inputs from set A and produces outputs in set B:

f: A → B

Then f⁻¹ reverses this — it takes inputs from B and produces outputs in A:

f⁻¹: B → A

Example

If f(x) = x², x ≥ 0

Domain of f: x ≥ 0

Range of f: y ≥ 0

Then for f⁻¹(x) = √x:

Domain of f⁻¹: x ≥ 0 ✓

Range of f⁻¹: y ≥ 0 ✓

For Quadratics

Always state the domain restriction. The range of the original function becomes the domain of the inverse.

Worked Example06

f(x) = x² − 6x + 5, x ≥ 3. Find f⁻¹(x) and state its domain and range.

1

Complete the square: f(x) = (x − 3)² − 9 + 5 = (x − 3)² − 4

2

Find range of f: When x ≥ 3, (x − 3)² ≥ 0, so f(x) ≥ −4. Range: y ≥ −4

3

Write y = (x − 3)² − 4, swap: x = (y − 3)² − 4

4

Rearrange: x + 4 = (y − 3)², so y − 3 = ±√(x + 4)

5

Since x ≥ 3 (right branch), use positive root: y = 3 + √(x + 4)

•

f⁻¹(x) = 3 + √(x + 4)

6

Domain of f⁻¹ = Range of f = x ≥ −4

7

Range of f⁻¹ = Domain of f = y ≥ 3

Worked Example07

Find the inverse of f(x) = (2x + 1)/(x − 3), x ≠ 3.

1

Write y = (2x + 1)/(x − 3), swap: x = (2y + 1)/(y − 3)

2

Multiply: x(y − 3) = 2y + 1

3

Expand: xy − 3x = 2y + 1

4

Collect y terms: xy − 2y = 3x + 1

5

Factorise: y(x − 2) = 3x + 1

6

Therefore: f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2

Graphical Relationship

Key Result

The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.

Why Reflection?

If (a, b) is on f, then f(a) = b

So f⁻¹(b) = a, meaning (b, a) is on f⁻¹

Example: (1, 3) on f ↔ (3, 1) on f⁻¹

Intersection

f and f⁻¹ always intersect on the line y = x

To find: solve f(x) = x

Finding Where f(x) = f⁻¹(x)

Key Insight

Since f and f⁻¹ intersect on the line y = x, to find where f(x) = f⁻¹(x), simply solve:

f (x) = x

Why This Works

At intersection points of f and f⁻¹, both equal the same y-value.

These points lie on y = x, so the y-value equals the x-value.

Therefore f(x) = x at these points.

Much Easier!

Instead of finding f⁻¹(x) then solving f(x) = f⁻¹(x) (complex algebra), just solve f(x) = x directly.

Worked Example08

Find the x-coordinate where f(x) = f⁻¹(x) for f(x) = 2x − 3.

1

To find where f(x) = f⁻¹(x), solve f(x) = x

2

2x − 3 = x

3

2x − x = 3

4

x = 3

•

Answer: x = 3

Verify: f(3) = 2(3) − 3 = 3, so the point is (3, 3) which lies on y = x ✓

Practice

Question 1: Given f(x) = x + 5 and g(x) = 2x − 1, find fg(x) and gf(x).

fg(x) = f(g(x)) = f(2x − 1) = (2x − 1) + 5 = 2x + 4

gf(x) = g(f(x)) = g(x + 5) = 2(x + 5) − 1 = 2x + 10 − 1 = 2x + 9

Practice

Question 2: Find the inverse of f(x) = 4x + 7.

y = 4x + 7

Swap: x = 4y + 7

Rearrange: x − 7 = 4y

y = (x − 7)/4

f⁻¹(x) = (x − 7)/4

Practice

Question 3: Find where f(x) = f⁻¹(x) for f(x) = 3x + 4.

Solve f(x) = x:

3x + 4 = x

3x − x = −4

2x = −4

x = −2

At x = −2, both f(x) and f⁻¹(x) equal −2.

Summary

Key Results

ConceptKey Point

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

Orderfg(x) ≠ gf(x) in general

Finding f⁻¹Swap x and y, rearrange for y

Domain/RangeDom(f⁻¹) = Range(f), Range(f⁻¹) = Dom(f)

Graphf⁻¹ is reflection of f in y = x

f(x) = f⁻¹(x)Solve f(x) = x (much easier!)

Exam Skills

Always check order carefully for composites

For quadratic inverses, note the domain restriction

To find f(x) = f⁻¹(x), just solve f(x) = x