
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)
1. $$f$$ continuous implies that $$f^{ -1}$$ is continuous.
2. $$f$$  increasing implies that  $$f^{ -1}$$ is increasing.
3. $$f$$ decreasing implies that $$f^{ -1}$$ is decreasing.
4. $$f$$ differentiable at $$c$$ and $$f '(c) \neq 0$$ implies that $$f^{ -1}$$ is differentiable at $$f (c)$$.
5. 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

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