Inverse Functions

IB Mathematics: Applications and Interpretation SL· Topic 2.4· 25 min read

1. One-to-One Functions and Invertibility★★☆☆☆⏱ 7 min

A function can only have a valid inverse if it is one-to-one (injective). A one-to-one function maps every output to exactly one input, so it must pass both the vertical line test (to qualify as a function) and the horizontal line test.

📘 Definition

One-to-one function

A function $f$ is one-to-one if whenever $f (a) = f (b)$ , it follows that $a = b$ . No two distinct inputs produce the same output.

Example:

$f (x) = 2 x + 3$ is one-to-one; $f (x) = x^{2}$ is not one-to-one over all real numbers.

📐 Worked Example

Prove that $f (x) = 3 x - 2$ is one-to-one over all real $x$ .

1
Assume $f (a) = f (b)$ for inputs $a$ and $b$ :
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$3a - 2 = 3b - 2$
3
Add 2 to both sides, then divide by 3:
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$3a = 3b \implies a = b$
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Since $f (a) = f (b)$ implies $a = b$ , $f (x)$ satisfies the definition of a one-to-one function.

Exam tip:

Always check for one-to-one status before finding an inverse; non one-to-one functions do not have valid inverses over their full domain.

2. Finding Inverses Algebraically★★★☆☆⏱ 8 min

Follow this standard 4-step process to find the inverse of any one-to-one function algebraically:

Start with the equation $y = f (x)$
Swap the positions of $x$ and $y$
Rearrange the equation to solve for $y$
The resulting $y$ is $f^{- 1} (x)$

📐 Worked Example

Find the inverse of $f (x) = \frac{2 x + 1}{x - 3}$ .

1
Step 1: Write the function with $y$ :
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$y = \frac{2x + 1}{x - 3}$
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Step 2: Swap $x$ and $y$ :
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$x = \frac{2y + 1}{y - 3}$
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Step 3: Multiply both sides by $(y - 3)$ and expand:
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$x(y - 3) = 2y + 1 \implies xy - 3x = 2y + 1$
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Step 4: Collect terms with $y$ on one side and factor:
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$xy - 2y = 3x + 1 \implies y(x - 2) = 3x + 1$
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Step 5: Solve for $y$ to get the inverse:
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$f^{-1}(x) = \frac{3x + 1}{x - 2}$

3. Graphical Properties of Inverse Functions★★☆☆☆⏱ 5 min

Swapping $x$ and $y$ for inverse functions means the graph of $f^{- 1} (x)$ is the reflection of the graph of $f (x)$ over the line $y = x$ . Any point $(a, b)$ on $f (x)$ maps to $(b, a)$ on $f^{- 1} (x)$ .

📐 Worked Example

Given $f (x) = 2 x + 1$ crosses the axes at $(0, 1)$ and $(- 0.5, 0)$ , sketch the inverse by reflection.

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Step 1: Swap the coordinates of the intercepts to get points on $f^{- 1} (x)$ :
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Original $(0, 1)$ → $(1, 0)$ ; original $(- 0.5, 0)$ → $(0, - 0.5)$
3
Step 2: Reflect all points of $f (x)$ over the line $y = x$ to get the full graph of $f^{- 1} (x)$ .
4
Step 3: Verify: the inverse is $f^{- 1} (x) = \frac{x - 1}{2}$ , which passes through both new points, confirming the reflection is correct.

4. Domain and Range of Inverse Functions★★★☆☆⏱ 5 min

Because inputs and outputs are swapped between a function and its inverse, there is a fixed relationship between their domains and ranges:

📘 Definition

Domain-Range Relationship

For any invertible function $f$ : $dom (f^{- 1}) = range (f)$ and $range (f^{- 1}) = dom (f)$

Example:

If $f (x) = x^{2}$ with domain $x \geq 0$ , range $y \geq 0$ , so $f^{- 1} (x) = x$ has domain $x \geq 0$ , range $y \geq 0$ .

📐 Worked Example

Given $f (x) = x^{2}$ with domain $x \geq 2$ and range $y \geq 4$ , find the domain and range of $f^{- 1} (x)$ .

1
Apply the domain-range relationship:
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1. Domain of $f^{- 1}$ = Range of original $f$
3
$\text{dom}(f^{-1}) = x \geq 4$
4
1. Range of $f^{- 1}$ = Domain of original $f$
5
$\text{range}(f^{-1}) = y \geq 2$
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This matches the inverse $f^{- 1} (x) = x$ , confirming the result.

5. Common Pitfalls

Wrong move:

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

Why:

The notation $f^{- 1}$ denotes inverse, not exponent -1.

Correct move:

Remember $f^{- 1} (x)$ undoes the action of $f (x)$ , it is not the reciprocal. For $f (x) = 2 x$ , $f^{- 1} (x) = \frac{x}{2}$ , not $\frac{1}{2 x}$ .

Wrong move:

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

Why:

Non one-to-one functions do not pass the horizontal line test, so their inverse would not be a valid function.

Correct move:

Always check for one-to-one status first, and restrict the domain of the original function to make it one-to-one if needed.

Wrong move:

Swapping the domain and range of the inverse function

Why:

It is easy to mix up which set belongs to which function.

Correct move:

Memorize: domain of inverse = range of original, range of inverse = domain of original to avoid this mistake.

Wrong move:

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

Why:

Common mistake when sketching inverse graphs from memory.

Correct move:

Always use $y = x$ as the mirror line, and swap the coordinates of 1-2 key points to confirm your reflection is correct.

6. Quick Reference Cheatsheet

Concept	Key Rule/Property
Inverse existence	Only for one-to-one functions (pass horizontal line test)
Algebraic steps	$y = f (x)$ 2. Swap $x, y$ 3. Solve for $y$ 4. $y = f^{- 1} (x)$
Graph property	$f^{- 1} (x)$ is reflection of $f (x)$ over $y = x$
Domain/Range	$dom (f^{- 1}) = range (f)$ , $range (f^{- 1}) = dom (f)$
Notation	$f^{- 1} (x) \neq = \frac{1}{f ( x )}$

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.

2021 · 1
Find inverse of restricted quadratic
2023 · 2
Graph inverse of a linear function

Going deeper

syllabusIB AI SL Topic 2 FunctionsCovers inverse function learning outcomes

What's Next

Inverse functions are a foundational concept for all future work with functions in IB AI SL. Most importantly, logarithms are defined as the inverse of exponential functions, so understanding inverses will make learning exponential and logarithmic models much simpler. Inverses are also commonly used to solve application problems where you need to reverse a function to find an input for a given output, which is a frequent requirement in exam questions. Mastering inverses will set you up for success for all upcoming function topics.