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Mathematics

Composite & Inverse Functions

PDF
Matthew Williams
|May 6, 2026|5 min read
CompositeFunctionsInversePaper 01Paper 02

Finding and evaluating composite functions, and how to find and verify an inverse.

Composite and inverse functions are about processes. A composite function performs one rule and then another; an inverse function reverses a rule and takes you back to the original input.

In CSEC, this topic sits in Relations, Functions and Graphs, where questions may ask you to evaluate, form, or rearrange functions. Pay close attention to order: f(g(x))f(g(x))f(g(x)) usually gives a different result from g(f(x))g(f(x))g(f(x)). For inverse functions, explain how each algebraic step undoes the original operation.

What Is a Composite Function?

A composite function is when you apply one function, then apply another function to the result.

Notation: (f∘g)(x)(f \circ g)(x)(f∘g)(x) or f(g(x))f(g(x))f(g(x))

Read as: "fff composed with ggg of xxx"

Process:

  1. First, find g(x)g(x)g(x) (apply inner function)
  2. Then, apply fff to that result: f(g(x))f(g(x))f(g(x))

Important: f(g(x))≠g(f(x))f(g(x)) \neq g(f(x))f(g(x))=g(f(x)) (order matters!)

Finding Composite Functions

Example

Given f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x2g(x) = x^2g(x)=x2, find f(g(x))f(g(x))f(g(x)):

Step 1: Identify what we're doing f(g(x))=apply g first, then apply ff(g(x)) = \text{apply } g \text{ first, then apply } ff(g(x))=apply g first, then apply f

Step 2: Find g(x)=x2g(x) = x^2g(x)=x2

Step 3: Apply fff to the result: wherever we see xxx in the definition of fff, substitute g(x)=x2g(x) = x^2g(x)=x2 f(g(x))=f(x2)=2(x2)+1=2x2+1f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1f(g(x))=f(x2)=2(x2)+1=2x2+1

Answer: f(g(x))=2x2+1f(g(x)) = 2x^2 + 1f(g(x))=2x2+1

This is a new function that takes xxx, squares it, multiplies by 2, and adds 1.

Example

Given f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x2g(x) = x^2g(x)=x2, find g(f(x))g(f(x))g(f(x)):

Step 1: Apply fff first to get f(x)=2x+1f(x) = 2x + 1f(x)=2x+1

Step 2: Apply ggg to the result: wherever xxx appears in ggg, substitute f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 g(f(x))=g(2x+1)=(2x+1)2g(f(x)) = g(2x+1) = (2x+1)^2g(f(x))=g(2x+1)=(2x+1)2

Step 3: Expand g(f(x))=4x2+4x+1g(f(x)) = 4x^2 + 4x + 1g(f(x))=4x2+4x+1

Compare:

  • f(g(x))=2x2+1f(g(x)) = 2x^2 + 1f(g(x))=2x2+1
  • g(f(x))=4x2+4x+1g(f(x)) = 4x^2 + 4x + 1g(f(x))=4x2+4x+1

They're DIFFERENT! Order matters!

Evaluating Composite Functions

To find f(g(3))f(g(3))f(g(3)) where f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x2g(x) = x^2g(x)=x2:

Method 1: Find g(x)g(x)g(x) first, then fff of that g(3)=32=9g(3) = 3^2 = 9g(3)=32=9 f(g(3))=f(9)=2(9)+1=19f(g(3)) = f(9) = 2(9) + 1 = 19f(g(3))=f(9)=2(9)+1=19

Method 2: Use the composite function formula we found f(g(x))=2x2+1f(g(x)) = 2x^2 + 1f(g(x))=2x2+1 f(g(3))=2(3)2+1=18+1=19f(g(3)) = 2(3)^2 + 1 = 18 + 1 = 19f(g(3))=2(3)2+1=18+1=19

Both methods give the same answer (as they should!).

Inverse Functions

What Is an Inverse Function?

An inverse function "undoes" what the original function does.

If fff takes input xxx to output yyy, then f−1f^{-1}f−1 takes input yyy back to output xxx.

Notation: f−1(x)f^{-1}(x)f−1(x) (read as "fff inverse of xxx")

Key property: f(f−1(x))=xf(f^{-1}(x)) = xf(f−1(x))=x f−1(f(x))=xf^{-1}(f(x)) = xf−1(f(x))=x

When Does an Inverse Exist?

Not every function has an inverse! A function has an inverse ONLY if it's one-to-one (injective):

  • Each output comes from exactly one input
  • No two different inputs produce the same output
  • Passes the horizontal line test: any horizontal line crosses the graph at most once
Example

Does f(x)=2x+3f(x) = 2x + 3f(x)=2x+3 have an inverse?

Check: Different inputs give different outputs

  • f(1)=5f(1) = 5f(1)=5
  • f(2)=7f(2) = 7f(2)=7
  • f(3)=9f(3) = 9f(3)=9
  • ...

Each input xxx produces a unique output. YES, it has an inverse. ✓

Does g(x)=x2g(x) = x^2g(x)=x2 have an inverse (for all real numbers)?

Check: Multiple inputs give same output

  • g(2)=4g(2) = 4g(2)=4
  • g(−2)=4g(-2) = 4g(−2)=4

Same output (4) from different inputs (2 and -2). NO, it doesn't have an inverse for all reals. ✗

(Note: If we restrict the domain to x≥0x \geq 0x≥0, then ggg is one-to-one and has an inverse.)

Finding Inverse Functions Algebraically

To find f−1(x)f^{-1}(x)f−1(x) from f(x)f(x)f(x):

Step 1: Write y=f(x)y = f(x)y=f(x)

Step 2: Swap xxx and yyy

Step 3: Solve for yyy in terms of xxx

Step 4: Replace yyy with f−1(x)f^{-1}(x)f−1(x)

Example

Find the inverse of f(x)=2x+3f(x) = 2x + 3f(x)=2x+3:

Step 1: Write y=2x+3y = 2x + 3y=2x+3

Step 2: Swap xxx and yyy x=2y+3x = 2y + 3x=2y+3

Step 3: Solve for yyy x−3=2yx - 3 = 2yx−3=2y y=x−32y = \frac{x-3}{2}y=2x−3​

Step 4: Write as inverse function f−1(x)=x−32f^{-1}(x) = \frac{x-3}{2}f−1(x)=2x−3​

Verify:

  • f(f−1(x))=f(x−32)=2⋅x−32+3=(x−3)+3=xf(f^{-1}(x)) = f\left(\frac{x-3}{2}\right) = 2 \cdot \frac{x-3}{2} + 3 = (x-3) + 3 = xf(f−1(x))=f(2x−3​)=2⋅2x−3​+3=(x−3)+3=x ✓
  • f−1(f(x))=f−1(2x+3)=(2x+3)−32=2x2=xf^{-1}(f(x)) = f^{-1}(2x+3) = \frac{(2x+3)-3}{2} = \frac{2x}{2} = xf−1(f(x))=f−1(2x+3)=2(2x+3)−3​=22x​=x ✓
Finding the Inverse of a Linear Function
Example

Find the inverse of f(x)=x3−2f(x) = x^3 - 2f(x)=x3−2:

Step 1: Write y=x3−2y = x^3 - 2y=x3−2

Step 2: Swap xxx and yyy x=y3−2x = y^3 - 2x=y3−2

Step 3: Solve for yyy x+2=y3x + 2 = y^3x+2=y3 y=x+23y = \sqrt[3]{x+2}y=3x+2​

Step 4: Write as inverse f−1(x)=x+23f^{-1}(x) = \sqrt[3]{x+2}f−1(x)=3x+2​

Verify: f(f−1(x))=(x+23)3−2=(x+2)−2=xf(f^{-1}(x)) = (\sqrt[3]{x+2})^3 - 2 = (x+2) - 2 = xf(f−1(x))=(3x+2​)3−2=(x+2)−2=x ✓

Inverse Functions and Composition

The fundamental relationship: f∘f−1=I (identity function)f \circ f^{-1} = I \text{ (identity function)}f∘f−1=I (identity function) f−1∘f=If^{-1} \circ f = If−1∘f=I

And if you have composite functions: (f∘g)−1=g−1∘f−1(f \circ g)^{-1} = g^{-1} \circ f^{-1}(f∘g)−1=g−1∘f−1

Note the order reverses when taking inverses of composites!

Example

If f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x−3g(x) = x - 3g(x)=x−3, find (f∘g)−1(f \circ g)^{-1}(f∘g)−1:

Method 1: Find (f∘g)(f \circ g)(f∘g) first, then invert

f(g(x))=f(x−3)=2(x−3)+1=2x−5f(g(x)) = f(x-3) = 2(x-3) + 1 = 2x - 5f(g(x))=f(x−3)=2(x−3)+1=2x−5

Now invert y=2x−5y = 2x - 5y=2x−5: x=2y−5x = 2y - 5x=2y−5 y=x+52y = \frac{x+5}{2}y=2x+5​

(f∘g)−1(x)=x+52(f \circ g)^{-1}(x) = \frac{x+5}{2}(f∘g)−1(x)=2x+5​

Method 2: Use (f∘g)−1=g−1∘f−1(f \circ g)^{-1} = g^{-1} \circ f^{-1}(f∘g)−1=g−1∘f−1

Find inverses:

  • f−1(x)=x−12f^{-1}(x) = \frac{x-1}{2}f−1(x)=2x−1​
  • g−1(x)=x+3g^{-1}(x) = x + 3g−1(x)=x+3

g−1(f−1(x))=g−1(x−12)=x−12+3=x−1+62=x+52g^{-1}(f^{-1}(x)) = g^{-1}\left(\frac{x-1}{2}\right) = \frac{x-1}{2} + 3 = \frac{x-1+6}{2} = \frac{x+5}{2}g−1(f−1(x))=g−1(2x−1​)=2x−1​+3=2x−1+6​=2x+5​

Both methods agree! ✓

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