Algebra and Trigonometry, 5th Edition

Inverse Functions and Composition

Suppose that we were to use some input a for a one-to-one function f and find its output, $f(a)$. The function $f^{-1}$ would then take that output back to a. Similarly, if we began with an input b for the function $f^{-1}$ and found its output, $f^{-1}(b)$, the original function f would then take that output back to b. This is summarized as follows.

If a function f is one-to-one, then $f^{-1}$ is the unique function such that each of the following holds:

$$\begin{array}{rcl}(f^{-1}\circ f)(x)=f^{-1}(f(x))& =& x,\mathrm{for\; each}x\text{ in the domain of }f,\text{ and}\\ (f\circ f^{-1})(x)=f(f^{-1}(x))& =& x,\mathrm{for\; each}x\text{ in the domain of }{f}^{-1}.\end{array}$$

Example 9

Given that $f(x)=5x+8$, use composition of functions to show that

$$f^{-1}(x)=\frac{x-8}{5}.$$

Solution

We find $(f^{-1}\circ f)(x)$ and $(f\circ f^{-1})(x)$ and check to see that each is ...