# 4.4: Inverse Functions

An inverse function is a function that undoes another function: If an input \(x\) into the function \(f\) produces an output \(y\), then putting \(y\) into the inverse function \(g\) produces the output \(x\), and vice versa.

Definition: Inverse Functions

Let \(f(x)\) be a 1-1 function then \(g(x)\) is an inverse function of \(f(x)\) if

\[ f(g(x)) = g(f(x)) = x. \]

Example 1

For

\[ f(x) = 2x - 1 \]

\[ f^{ -1}(x) = \dfrac{1}{2} x +\dfrac{1}{2} \]

since

\[ f(f^{ -1}(x) ) = 2[\dfrac{1}{2} x +\dfrac{1}{2}] - 1 = x \]

and

\[ f ^{-1}(f(x)) = \dfrac{1}{2} [2x - 1] + \dfrac{1}{2} = x. \]

### The Horizontal Line Test and Roll's Theorem

**Note** that if \(f(x)\) is differentiable and the horizontal line test fails then

\[ f(a) = f(b) \]

and *Rolls theorem* implies that there is a \(c\) such that

\[ f '(c) = 0.\]

A partial converse is also true:

## Theorem (Roll's Theorem) |

If \(f\) is differentiable and \(f '(x)\) is always non-negative (or always non-positive) then \(f(x)\) has an inverse. |

Example 2

\[ f(x) = x^3 + x - 4 \]

has an inverse since

\[f'(x) = 3x^2 + 1 \]

which is always positive.

### Continuity and Differentiability of the Inverse Function

## Theorem (Continuity and Differentiability) |

- \(f\) continuous implies that \(f^{ -1}\) is continuous.
- \(f\) increasing implies that \(f^{ -1}\) is increasing.
- \(f\) decreasing implies that \(f^{ -1}\) is decreasing.
- \(f\) differentiable at \(c\) and \(f '(c) \neq 0\) implies that \(f^{ -1}\) is differentiable at \(f (c)\).
- If \(g(x)\) is the inverse of the differentiable \(f(x)\) then
\[ g'(x) = \dfrac{1}{f '(g(x))}.\] if \(f '(g(x)) \neq 0\). |

**Proof of (5)**

Since

\[ f (g(x)) = x \]

we differentiate implicitly:

\[\dfrac{d}{dx} f(g(x)) = \dfrac{d}{dx} x.\]

Using the chain rule

\[ y =f(u), u = g(x)\]

\[ \dfrac{dy}{x} = \dfrac{dy}{dy} \dfrac{du}{dx} \]

\[ = f '(u) g'(x) = f '(g(x)) g'(x).\]

So that

\[ f '(g(x)) g'(x) = 1. \]

Dividing, we get:

\[g'(x) = \dfrac{1}{f'(g(x))}.\]

Example 3

For \(x > 0\), let

\[ f(x) = x^2\]

and

\[ g(x) = \sqrt{x}\]

be its inverse, then

\[ g'(x) = \dfrac{1}{2\sqrt{x}}.\]

Note that

\[ \dfrac{d}{dx} \sqrt{x} = \dfrac{d}{dx} x^{\frac{1}{2}} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}.\]

Exercise

1. Let

\[ f(x) = x^3 + x - 4. \]

Find

\[ \dfrac{d}{dx} f^{-1}(-4). \]

2. Let

\[f(x) =\int_2^x \dfrac{1}{1+x^3} dx\]

Find

\[ \dfrac{d}{dx} f^{-1}(0). \]

### Outside Links

### Contributors

- Larry Green (Lake Tahoe Community College)

Integrated by Justin Marshall.