Chapter 1.5, Problem 59E

To determine

To prove: If $f\left(x\right)$ is one-to-one than $g\left(x\right)=-f\left(x\right)$ is one-to-one.

Explanation of Solution

Given information: Given that $f$ is one-to-one.

Proof:

A function is said to be one-to-one if $f\left(a\right)=f\left(b\right)$ such that $a=b$ for all $a$ and $b$ in the domain of $f\left(x\right)$ .

It is required to prove that $g\left(a\right)=g\left(b\right)$ where $a=b$ for all $a$ and $b$ in the domain of $g\left(x\right)$ .

Let $a$ and $b$ be two values in the domain of $g\left(x\right)$ . Then,

  $\begin{array}{c}g\left(a\right)=g\left(b\right)\left\{\therefore g\left(x\right)=-f\left(x\right)\right\}\\ -f\left(a\right)=-f\left(b\right)\\ f\left(a\right)=f\left(b\right)\end{array}$

It is given that $f$ is a one-to-one function, $f\left(a\right)=f\left(b\right)$ where $a=b$ .

Hence, it is proved that $g\left(x\right)$ is also a one-to-one function.