Inverse Functions Inverse Functions

Introduction

Below is the function,

f
, where
f(A) = 3
,
f(C) = 4
, etc.

Notice how every input has an output. This is required for all functions. Also notice how some inputs have the same output.
A function is bijective if it follows the following rules:
1. Every input has a unique output. No two inputs share an output.
2. The function maps to every output. No output is left unmapped.


Below is an example of a bijective function,
g
. There are no extra outputs without mappings, and there are no shared outputs.
Bijective functions are special because they can be reversed. If given
g(x)=7
, we could deduce that
x=C
.

This reverse function, applying a function backwards, is called the inverse function.

Below is the inverse function of
g
:
g^{-1}
, where
g^{-1}(1)=B
,
g^{-1}(4)=G
,
g^{-1}(5)=D
For a bijective function,
f: X \to Y
, its inverse is
f^{-1}: Y \to X
, where if
f(x) = y
, then
f^{-1}(y) = x
.

As a result, applying the inverse function to a function reverses its effect.
f^{-1}(f(x))=x

Also, not only is
f^{-1}
the inverse of
f
,
f
is the inverse of
f^{-1}
.
f(f^{-1}(y))=y

Proof of the Derivative of the Inverse Function

Let

f
be a bijective function with an inverse function
f^{-1}
, where
f(x) = y
and
f(a) = b
.
f'(a)=\lim_{x \to a}\frac{f(x) - f(a)}{x - a}
Define the derivative of the inverse function.
(f^{-1})'(b)=\lim_{y \to b}\frac{f^{-1}(y) - f^{-1}(b)}{y - b}
Substitute
y
with
f(x)
and
b
with
f(a)
.
(f^{-1})'(b)=\lim_{f(x) \to f(a)}\frac{f^{-1}(f(x)) - f^{-1}(f(a))}{f(x) - f(a)}
Remove
f^{-1}(f)
.
(f^{-1})'(b)=\lim_{x \to a}\frac{x - a}{f(x) - f(a)}
Simplify.
(f^{-1})'(b)=\frac{1}{f'(a)}
Put in terms of
b
.
(f^{-1})'(b)=\frac{1}{f'(f^{-1}(b))}

Application

There are two definitions of the derivative of the inverse function, one using

f'(a)
and the other using
f'(f^{-1}(b))
. We will use both.

For example, let
f(x) = x^3 + x
. This function is bijective, so it has an inverse function. (We will prove that it is bijective in the next lesson.)

Find the derivative of
f^{-1}(y)
at
y=2
.
(f^{-1})'(b)=\frac{1}{f'(f^{-1}(b))}
(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}
f^{-1}(2)
is the value of
x
when
f(x)=2
.
\begin{align*} x^3 + x &= 2 \\\\ x&=1 \\\\ f(1)&=2 \\\\ f^{-1}(2)&=1 \\\\ (f^{-1})'(2)&=\frac{1}{f'(1)} \\\\ f'(x)&=3x^2 + 1 \\\\ f'(1)&=3(1)^2 + 1 \\\\ f'(1)&=4 \\\\ (f^{-1})'(2)&=\frac{1}{4} \end{align*}
The derivative of
f^{-1}(y)
at
y=2
is
1/4
.