Finding Inverse Functions and Their Graphs
Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\left(x\right)={x}^{2}[/latex] restricted to the domain [latex]\left[0,\infty \right)[/latex], on which this function is one-to-one, and graph it as below.

Restricting the domain to [latex]\left[0,\infty \right)[/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.
We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\left(x\right)=\sqrt{x}[/latex]. What happens if we graph both [latex]f\text{ }[/latex] and [latex]{f}^{-1}[/latex] on the same set of axes, using the [latex]x\text{-}[/latex] axis for the input to both [latex]f\text{ and }{f}^{-1}?[/latex]
We notice a distinct relationship: The graph of [latex]{f}^{-1}\left(x\right)[/latex] is the graph of [latex]f\left(x\right)[/latex] reflected about the diagonal line [latex]y=x[/latex], which we will call the identity line, shown below.

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.

Yes. If [latex]f={f}^{-1}[/latex], then [latex]f\left(f\left(x\right)\right)=x[/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because
[latex]\frac{1}{\frac{1}{x}}=x[/latex]
Any function [latex]f\left(x\right)=c-x[/latex], where [latex]c[/latex] is a constant, is also equal to its own inverse.