Composite and inverse functions

IB Mathematics: Analysis and Approaches HL· 2.5 Composite and inverse functions· 6 min read

1. Composite Functions★★☆☆☆⏱ 15 min

📘 Definition

Composite Function

(f \circ g)(x) = f(g(x))

A function where the output of the inner function $g (x)$ is used as the input to the outer function $f (x)$ .

Example:

If $f (x) = x^{2}$ and $g (x) = x + 1$ , then $(f \circ g) (x) = (x + 1)^{2}$

📐 Worked Example

Given $f (x) = 2 x + 3$ and $g (x) = e^{x - 1}$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$

1
For $(f \circ g) (x)$ , substitute $g (x)$ into $f (x)$ as the input:
2
$f(g(x)) = 2\left(g(x)\right) + 3 = 2e^{x-1} + 3$
3
For $(g \circ f) (x)$ , substitute $f (x)$ into $g (x)$ as the input:
4
$g(f(x)) = e^{f(x) - 1} = e^{(2x + 3) - 1} = e^{2x + 2}$

2. Domain and Range of Composite Functions★★★☆☆⏱ 20 min

To find the domain of a composite function $f (g (x))$ , you must apply two layers of restrictions: first $x$ must be valid for the inner function $g$ , and second the output of $g$ must be valid for the outer function $f$ .

Find the full domain of the inner function $g (x)$
Add additional restrictions to ensure all outputs of $g (x)$ are in the domain of $f (x)$
The domain of $f (g (x))$ is the intersection of these two sets of $x$ -values

📐 Worked Example

Find the domain of $(f \circ g) (x)$ where $f (x) = \frac{1}{x}$ and $g (x) = x + 2$

1
First find the domain of the inner function $g (x) = x + 2$ :
2
$x + 2 \geq 0 \implies x \geq -2, \quad D(g) = [-2, \infty)$
3
Next find restrictions from $f (x) = \frac{1}{x}$ , which is undefined at $x = 0$ , so $g (x) \neq = 0$ :
4
$\sqrt{x+2} \neq 0 \implies x+2 \neq 0 \implies x \neq -2$
5
Combine the two restrictions to get the final domain:
6
$D(f \circ g) = (-2, \infty)$

3. Finding Inverse Functions Algebraically★★★☆☆⏱ 20 min

📘 Definition

Inverse Function

f^{-1}(x)

A function that reverses the mapping of an original one-to-one function, satisfying $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ for all valid $x$ . Only exists if the original function is one-to-one.

Start with $y = f (x)$
Swap the variables $x$ and $y$
Rearrange the equation to solve for $y$ in terms of $x$
The resulting $y$ is $f^{- 1} (x)$ , state its domain (equal to the range of original $f$ )

📐 Worked Example

Find the inverse of $f (x) = 3 x + 4$ , $x \in R$

1
Write the original function:
2
$y = 3x + 4$
3
Swap $x$ and $y$ :
4
$x = 3y + 4$
5
Rearrange to solve for $y$ :
6
$x - 4 = 3y \implies y = \frac{x - 4}{3}$
7
Write as the inverse function:
8
$f^{-1}(x) = \frac{x - 4}{3}, \quad x \in \mathbb{R}$

✓ Quick check

Test your understanding of inverse notation:

Which of the following is the inverse of $f (x) = \frac{1}{x + 3}$ , $x \neq = - 3$ ?
- $\frac{1}{x} - 3$
- $\frac{1}{x} + 3$
- $x - 3$
- $\frac{1}{1 - 3 x}$

4. Graphical Properties of Inverses★★☆☆☆⏱ 15 min

Because we swap $x$ and $y$ when finding an inverse, every point $(a, b)$ on the original function $f$ becomes the point $(b, a)$ on the inverse function $f^{- 1}$ . This corresponds to a reflection of the entire graph of $f$ over the line $y = x$ .

📐 Worked Example

Sketch the graph of $f (x) = e^{x}$ and its inverse, using reflection rules

1
Original $f (x) = e^{x}$ has a y-intercept at $(0, 1)$ , horizontal asymptote at $y = 0$ , and is increasing for all $x \in R$
2
Reflect all points over the line $y = x$ , so the intercept $(0, 1)$ becomes $(1, 0)$ , and the horizontal asymptote $y = 0$ becomes the vertical asymptote $x = 0$
3
The reflected graph is the natural logarithm function, which matches the known inverse relationship:
4
$f^{-1}(x) = \ln x, \quad x > 0$

5. Common Pitfalls

Wrong move:

Calculating $g (f (x))$ instead of $f (g (x))$ for $f \circ g$

Why:

Notation is read right-to-left for application: the inner function closest to $x$ is applied first

Correct move:

Remember $f \circ g (x) = f (g (x))$ , so $g$ is applied first, then $f$

Wrong move:

Confusing $f^{- 1} (x)$ with $\frac{1}{f ( x )}$

Why:

The negative exponent notation for inverse functions is a special convention, not a reciprocal

Correct move:

Always remember $f^{- 1} (x)$ means inverse function, not reciprocal: $(f (x))^{- 1} = \frac{1}{f ( x )} \neq = f^{- 1} (x)$

Wrong move:

Finding an inverse for a non-one-to-one function without domain restriction

Why:

A non-one-to-one function's reflection will not pass the vertical line test, so it is not a valid function

Correct move:

Always check if $f$ is one-to-one first, and restrict its domain to make it one-to-one if required

Wrong move:

Only taking the domain of the outer function when finding the domain of a composite

Why:

The input $x$ must first be valid for the inner function, so all restrictions on the inner function apply

Correct move:

Start by finding the domain of the inner function, then add restrictions from the outer function

Wrong move:

Reflecting over the x-axis instead of $y = x$ for inverse graphs

Why:

Reflection over x-axis gives $- f (x)$ , not the inverse, from misremembering the rule

Correct move:

Always reflect over the diagonal line $y = x$ to get the graph of $f^{- 1} (x)$

6. Quick Reference Cheatsheet

Concept	Key Rule	Notation
Composite $f \circ g$	Apply $g$ first, then $f$	$(f \circ g) (x) = f (g (x))$
Domain of $f (g (x))$	Domain of $g$ + output of $g$ in domain of $f$	$D (f \circ g) = D (g) \cap {x : g (x) \in D (f)}$
Inverse existence	Only exists for one-to-one functions	$f^{- 1} (x)$
Find inverse algebraically	$y = f (x)$ 2. swap $x, y$ 3. solve for $y$	Domain of $f^{- 1}$ = range of $f$
Graph of $f^{- 1}$	Reflection of $f$ over $y = x$	$(a, b) \to (b, a)$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
6 mark composite domain question
2022 · 2
5 mark inverse graph sketch question
2021 · 1
4 mark inverse verification question

Going deeper

IB SyllabusIB AA HL Syllabus Topic 2Official syllabus reference

What's Next

Composite and inverse functions are foundational to almost all advanced topics in IB AA HL. Composition is the core of the chain rule for differentiation, one of the most heavily tested calculus concepts in the exam. Inverse functions are critical for understanding logarithms (the inverse of exponentials) and for integration, where you will often need to invert functions to solve for antiderivatives. Mastering the domain rules here will prevent simple errors on much higher-mark questions later in the course.