Composition of functions and inverse functions

IB Mathematics: Applications and Interpretation HL· Unit 2: Functions, Topic 2.2· 25 min read

1. Composition of Functions★★☆☆☆⏱ 8 min

📘 Definition

Composite function

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

A function formed by applying the output of inner function $g$ as the input to outer function $f$ , read as "f of g of x".

Example:

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

Order matters for composition: $f \circ g$ is almost never equal to $g \circ f$ . The domain of $f \circ g$ is all $x$ in the domain of $g$ where $g (x)$ is also in the domain of $f$ .

📐 Worked Example

Let $f (x) = x - 2$ and $g (x) = x^{2} - 5$ . Find $(f \circ g) (x)$ and state its domain.

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1. Substitute $g (x)$ into $f$ to get the composite expression:
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$(f \circ g)(x) = f(g(x)) = f(x^2 - 5) = \sqrt{(x^2 - 5) - 2} = \sqrt{x^2 - 7}$
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1. Find domain restrictions: $g (x)$ is defined for all real $x$ , so we only require the expression under the square root to be non-negative:
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$x^2 - 7 \geq 0 \implies x^2 \geq 7 \implies x \leq -\sqrt{7} \text{ or } x \geq \sqrt{7}$
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1. Final domain in interval notation: $(- \infty, - 7] \cup [7, \infty)$

Exam tip:

Always check domain restrictions from both the inner and outer functions when calculating composite domains.

2. One-to-One Functions and Inverse Functions★★★☆☆⏱ 8 min

📘 Definition

Inverse function

$f^{- 1} (x)$

A function that reverses the mapping of a one-to-one function $f$ , satisfying $f^{- 1} (f (x)) = x$ for all $x$ in the domain of $f$ , and $f (f^{- 1} (x)) = x$ for all $x$ in the domain of $f^{- 1}$ . Only one-to-one functions have inverse functions.

Example:

If $f (x) = 2 x + 4$ , then $f^{- 1} (x) = \frac{x - 4}{2}$

A function is one-to-one if it passes the horizontal line test: no horizontal line intersects its graph more than once. If a function is not one-to-one over its natural domain, you can restrict the domain to create a one-to-one function that has an inverse.

📐 Worked Example

Find the inverse of $f (x) = 3 x - 6$ and verify it is correct.

1
1. Replace $f (x)$ with $y$ :
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$y = 3x - 6$
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1. Swap $x$ and $y$ , a required step for finding inverses:
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$x = 3y - 6$
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1. Solve for $y$ :
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$x + 6 = 3y \implies y = \frac{x + 6}{3} = \frac{1}{3}x + 2$
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1. Replace $y$ with $f^{- 1} (x)$ :
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$f^{-1}(x) = \frac{1}{3}x + 2$
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1. Verify by checking $f (f^{- 1} (x)) = x$ :
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$f\left(\frac{1}{3}x + 2\right) = 3\left(\frac{1}{3}x + 2\right) - 6 = x + 6 - 6 = x$
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The inverse is confirmed correct.

3. Graphical Properties of Inverses★★☆☆☆⏱ 5 min

The most important graphical property of inverse functions is the reflection rule: the graph of $y = f^{- 1} (x)$ is the reflection of the graph of $y = f (x)$ across the line $y = x$ . This comes directly from swapping $x$ and $y$ when finding the inverse.

This reflection rule also gives a useful domain/range relationship: the domain of the original function is the range of its inverse, and the range of the original function is the domain of its inverse.

📐 Worked Example

The function $f (x) = (x - 2)^{2}$ is restricted to the domain $x \geq 2$ , range $y \geq 0$ . What is the domain and range of $f^{- 1} (x)$ ?

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1. Use the domain/range swap rule for inverses:
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Domain of $f^{- 1} (x)$ = Range of $f (x)$ = $[0, \infty)$
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Range of $f^{- 1} (x)$ = Domain of $f (x)$ = $[2, \infty)$
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Graphically, the inverse is a reflection of the restricted parabola across $y = x$ , which matches this result.

4. Composition and Inverses in Context★★★☆☆⏱ 4 min

In IB AI HL exams, you will often encounter these concepts in real-world contexts. Composition combines two sequential relationships, while inverses reverse a relationship to find an input from a given output.

📐 Worked Example

The temperature in Celsius is converted to Fahrenheit with $F (C) = \frac{9}{5} C + 32$ . Find the inverse function to convert Fahrenheit back to Celsius, then find the Celsius equivalent of 77°F.

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1. Write the original function, swap $F$ and $C$ :
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$F = \frac{9}{5}C + 32 \implies C = \frac{9}{5}F^{-1}(C) + 32$
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1. Solve for the inverse function $F^{- 1} (C)$ :
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$C - 32 = \frac{9}{5}F^{-1}(C) \implies F^{-1}(C) = \frac{5}{9}(C - 32)$
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1. Substitute 77°F into the inverse:
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$F^{-1}(77) = \frac{5}{9}(77 - 32) = \frac{5}{9}(45) = 25^\circ C$

5. Common Pitfalls

Wrong move:

Confusing $(f \circ g) (x)$ with $f (x) \times g (x)$

Why:

The open circle notation for composition is not a multiplication symbol

Correct move:

Remember $(f \circ g) (x)$ means substitute $g (x)$ into $f (x)$ as input, i.e. $f (g (x))$

Wrong move:

Treating $f^{- 1} (x)$ as the reciprocal $\frac{1}{f ( x )}$

Why:

The negative exponent notation for inverses does not follow the usual reciprocal rule

Correct move:

$f^{- 1} (x)$ always refers to the inverse function that reverses the mapping of $f (x)$ , never the reciprocal

Wrong move:

Only checking domain restrictions from the outer function for composite functions

Why:

The inner function may already exclude inputs that are valid for the outer function

Correct move:

First find the domain of the inner function, then apply any additional restrictions from the outer function

Wrong move:

Finding an inverse for a many-to-one function without restricting the domain

Why:

A many-to-one function fails the horizontal line test, so it does not have a valid inverse function over its full domain

Correct move:

Restrict the domain of the original function to an interval where it is one-to-one before finding the inverse

6. Quick Reference Cheatsheet

Concept	Key Rule	Notation
Composition $f$ after $g$	Substitute $g (x)$ into $f (x)$	$(f \circ g) (x) = f (g (x))$
Domain of $f \circ g$	x in domain(g), g(x) in domain(f)
Inverse function	Reverses mapping of one-to-one $f$	$f^{- 1} (x) \neq = \frac{1}{f ( x )}$
Inverse verification	$f (f^{- 1} (x)) = x$ and $f^{- 1} (f (x)) = x$
Graph of inverse	Reflection over $y = x$ from $y = f (x)$
Domain/range swap	Domain( $f$ ) = Range( $f^{- 1}$ )

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.

2025 · 1
Find inverse of restricted quadratic
2024 · 2
Composite function domain problem
2023 · 1
Verify inverse function pair

What's Next

Composition and inverse functions are foundational for almost all remaining function topics in IB AI HL. Transformations of functions can be described as compositions of the original function with linear shifting and scaling functions. Exponential and logarithmic functions are inverses of each other, so understanding their inverse relationship is key to solving equations with these functions. In statistics, inverse functions are used to calculate critical values and inverse normal probabilities, which are a core exam topic.