Math Interactive
2. Functions
2.2. Inverse functions
2.2.1. Existence
Important
Motivating example

Recall that being a function means that every x{x}-value corresponds to exactly one f(x){f(x)}-value.

Consider the example f(x)=x2{f(x)=x^2} for x∈R{x \in \mathbb{R}}. It is a function and satisfies the above condition. In particular, x=−2{x=-2} corresponds to exactly one f(x){f(x)}-value 4{4}. Similarly, f(2)=4{f(2)=4}.

In trying to set up the inverse, however, we run into a problem. The value of 4{4} would correspond to both −2{-2} and 2{2}, leaving us with more than one possible output for the inverse relation. Thus the inverse is not a function, and we say that f−1{f^{-1}} does not exist.

For f−1{f^{-1}} to exist, we then need each f(x){f(x)}-value in the range of f{f} to correspond to a unique x{x}-value in the domain. We call such functions one-to-one.

One-to-one functions

Given a function f,f, its inverse function f−1f^{-1} exists if ff is one-to-one.

Horizontal line test

Determining whether a function is one-to-one (and hence whether it has an inverse) can be done with the horizontal line test.

A function ff is one-to-one if all horizontal lines y=k,k∈Ry=k, k \in \mathbb{R} cut the graph of y=f(x)y=f(x) at most once.

A function ff is not one-to-one if there is a horizontal line y=k0y=k_0 that cuts the graph of y=f(x)y=f(x) more than once.

x{x}
y{y}
(−1,10){\left(-1 , 10 \right)}
(13,−23){\left(\frac{1}{3} , - \frac{2}{3} \right)}
f:x↦6x2−4xfor x∈R,x≥−1.{f: x \mapsto 6 x^2 - 4 x} \allowbreak \quad \allowbreak {\textrm{for } x \in \mathbb{R}, x \geq -1.}
The horizontal line y=5{y=5} cuts the graph of y=f(x){y=f(x)} more than once. Hence f{f} is not a one-to-one function and the inverse function f−1{f^{-1}} does not exist.
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