Chapter 1.5, Problem 60E

To determine

To prove:The function $g\left(x\right)=\frac{1}{f\left(x\right)}$ is one-to-one if $f$ is a one-to-one , and $f\left(x\right)$ is never zero.

Explanation of Solution

Given information:The function $f$ is a one-to-one , and $f\left(x\right)$ is never zero.

Proof:

Let $a$ and $b$ be any two values in the domain of $g\left(x\right)$ such that $g\left(a\right)=g\left(b\right)$ .

Since $g\left(x\right)=\frac{1}{f\left(x\right)}$ , then $g\left(a\right)=g\left(b\right)$ . Now,

  $\begin{array}{c}\frac{1}{f\left(a\right)}=\frac{1}{f\left(b\right)}\\ f\left(a\right)=f\left(b\right)\end{array}$

Since, $f$ is a one-to-one function, and $f\left(a\right)=f\left(b\right)$ implies $a=b$ .

Now,

  $\begin{array}{c}g\left(a\right)=g\left(b\right)\\ \frac{1}{f\left(a\right)}=\frac{1}{f\left(b\right)}\\ f\left(a\right)=f\left(b\right)\\ a=b\end{array}$

Hence, it is prove that $g\left(x\right)=\frac{1}{f\left(x\right)}$ is one-to-one.