Formula of inverse functions

Adapted from OpenStax Calculus Volume 1¹

After confirming that an inverse function exists, we now attempt to find the formula that defines the inverse function.

Finding a function’s inverse

Let $y = f (x) .$
Make $x$ the subject of the equation in terms of $y .$
Interchange the variables $x$ and $y$ and write $y = f^{- 1} (x) .$

Example

Simpler examples

Use the technique above to find the formula for the inverses of $f (x) = 3 x - 4$ and $g (x) = 2 e^{x + 1} - 5 .$

You should get $f^{- 1} (x) = \frac{x + 4}{3}$ and $g^{- 1} (x) = \ln (\frac{x + 5}{2}) - 1 .$

Inverse of a quadratic function

Some tricks can help us find the inverse of a quadratic function. First, completing the square may be useful to help us make $x$ the subject.

We will then have a $\pm$ when taking square roots, and will need to use the domain to determine if we should take the positive or negative version of our expression.

The following example illustrates both of these concepts:

Question:
The function $f$ is given by

$f : x \mapsto x^{2} + 4 x, x \in ℝ, x \leq - 2,$

Define $f^{- 1}$ in similar form.

Solution:

$\begin{aligned} y = x^{2} + 4 x \\ y = (x + 2)^{2} - 4 \\ (x + 2)^{2} = y + 4 \\ x + 2 = \pm \sqrt{y + 4} \end{aligned}$

Since $x \leq - 2$ (the domain of $f$ ),

$\begin{aligned} x + 2 & = - \sqrt{y + 4} \\ x & = - 2 - \sqrt{y + 4} \\ f^{- 1} (x) & = - 2 - \sqrt{x + 4} \end{aligned}$

Note that $D_{f^{- 1}} = R_{f} = [- 4, \infty) .$
Hence the definition of $f^{- 1}$ is

$f^{- 1} : x \mapsto - 2 - \sqrt{x + 4}, x \in ℝ, x \geq - 4 . ∎$

Content in this page is adapted from OpenStax Calculus Volume 1 by Gilbert Strang and Edwin “Jed” Herman under the Creative Commons Attribution Noncommercial Sharealike 4.0 License.
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-4-inverse-functions ↩︎