AP®MathsCollege BoardPrecalculusRevision NotesExponential & Logarithmic FunctionsComposite & Inverse FunctionsDetermining an Inverse Function

Determining an Inverse Function (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 13 April 2026

Range & domain of an inverse function

When a function $f$ has an inverse $f^{- 1}$ , the inputs and outputs swap roles
This means:
- The domain of $f$ becomes the range of $f^{- 1}$
- The range of $f$ becomes the domain of $f^{- 1}$
E.g. if $f$ has domain $[0, 5]$ and range $[2, 10]$
- then $f^{- 1}$ has domain $[2, 10]$ and range $[0, 5]$
This relationship also applies to values in tables
- If you want to find the inverse by swapping input-output pairs
  - then the column of input values ( $x$ -values) in the original table becomes the column of output values in the inverse table
  - and vice versa
- The point $(a, b)$ in $f$ corresponds to the point $(b, a)$ in $f^{- 1}$

Are there additional restrictions on the applicability of an inverse function?

Beyond the mathematical domain and range swap, contextual restrictions may also limit when an inverse function is meaningful
E.g. if $f (t)$ models the height of a ball as a function of time
- the inverse $f^{- 1} (h)$ gives the time at which the ball reaches height $h$
  - But this inverse may only be useful for heights the ball actually reaches
  - and only for the time interval during which the ball is in the air

Examiner Tips and Tricks

When applying inverse functions in real-world problems, always check whether the inputs and outputs make sense in the given context.

Finding an inverse function analytically

How do you find the inverse of a function from its formula?

To find the inverse of a function given by an equation:
- Start with the equation as $y = f (x)$
- Then swap $x$ and $y$
  - this gives the equation as $x = f (y)$
- Finally solve the resulting equation for $y$
  - The solution gives you $y = f^{- 1} (x)$
This method works by reversing the operations that $f$ performs on its input
E.g. find the inverse of $f (x) = 3 x - 5$ :
- Write
  - $y = 3 x - 5$
- Swap
  - $x = 3 y - 5$
- Solve
  - $x + 5 = 3 y$ , so $y = \frac{x + 5}{3}$
- Therefore $f^{- 1} (x) = \frac{x + 5}{3}$
E.g. find the inverse of $f (x) = 2 x^{3} - 5$ :
- Write
  - $y = 2 x^{3} - 5$
- Swap
  - $x = 2 y^{3} - 5$
- Solve
  - $x + 5 = 2 y^{3}$ , then $y^{3} = \frac{x + 5}{2}$ , so $y = \sqrt[3]{\frac{x + 5}{2}}$
- Therefore $f^{- 1} (x) = \sqrt[3]{\frac{x + 5}{2}}$

Composition of inverse functions

What happens when you compose a function with its inverse?

As $f$ and $f^{- 1}$ are inverses of each other, then composing them in either order returns the original input
- $f (f^{- 1} (x)) = x$
- $f^{- 1} (f (x)) = x$
In other words, applying a function and then its inverse (or vice versa) "undoes" the effect
- you end up back where you started
- $f$ and $f^{- 1}$ 'cancel each other'
The composition of a function with its inverse (in either order) is equivalent to the identity function
- I.e. the composition simply returns its input unchanged
This property is useful for verifying a proposed inverse
- Substitute one function into the other and check that the result simplifies to $x$
- E.g. for $f (x) = 3 x - 5$ and $f^{- 1} (x) = \frac{x + 5}{3}$ :
  - $f (f^{- 1} (x)) = 3 \cdot \frac{x + 5}{3} - 5 = (x + 5) - 5 = x$ ✓
  - $f^{- 1} (f (x)) = \frac{(3 x - 5) + 5}{3} = \frac{3 x}{3} = x$ ✓

Examiner Tips and Tricks

When finding the inverse from a formula in a free response question, show every step of the "swap and solve" process . On the exam, supporting work is expected.

To verify an inverse, you only need to show the composition in one direction (either $f (f^{- 1} (x)) = x$ or $f^{- 1} (f (x)) = x$ ), though showing both is fine.

Worked Example

The function $f$ is given by $f (x) = \frac{4 x + 7}{2}$ .

(a) Find $f^{- 1} (x)$ .

Answer:

Write as $y = f (x)$

$y = \frac{4 x + 7}{2}$

Swap $x$ and $y$

$x = \frac{4 y + 7}{2}$

Solve for $y$

$2 x = 4 y + 7$

$4 y = 2 x - 7$

$y = \frac{2 x - 7}{4}$

Therefore

$f^{- 1} (x) = \frac{2 x - 7}{4}$

(b) State the domain and range of $f^{- 1}$ , given that $f$ has domain $[- 3, 5]$ .

Answer:

The domain of $f$ is $[- 3, 5]$

You need the range of $f$ on this domain
$f$ is an increasing linear function, so you just need to worry about the endpoints

$f (- 3) = \frac{4 (- 3) + 7}{2} = \frac{- 5}{2} = - 2.5$

$f (5) = \frac{4 (5) + 7}{2} = \frac{27}{2} = 13.5$

range of $f$ is $[- 2.5, 13.5]$

The range and domain are swapped to get the range and domain for $f^{- 1}$

The domain of $f^{- 1}$ is $[- 2.5, 13.5]$ and the range of $f^{- 1}$ is $[- 3, 5]$

Answer:

Substitute the expression for $f^{- 1} (x)$ everywhere that $x$ appears in the expression for $f (x)$

$f (f^{- 1} (x)) = f (\frac{2 x - 7}{4}) = \frac{4 \cdot \frac{2 x - 7}{4} + 7}{2} = \frac{(2 x - 7) + 7}{2} = \frac{2 x}{2} = x$

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Determining an Inverse Function (College Board AP® Precalculus): Revision Note

Range & domain of an inverse function

Are there additional restrictions on the applicability of an inverse function?

Examiner Tips and Tricks

Finding an inverse function analytically

How do you find the inverse of a function from its formula?

Composition of inverse functions

What happens when you compose a function with its inverse?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis

Determining an Inverse Function (College Board AP® Precalculus): Revision Note

Range & domain of an inverse function

How are the domain and range of a function and its inverse related?

Are there additional restrictions on the applicability of an inverse function?

Finding an inverse function analytically

How do you find the inverse of a function from its formula?

Composition of inverse functions

What happens when you compose a function with its inverse?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis