Chapter 1.6, Problem 90E

To determine

To find : the domain of the function so that the function is one-to-one and has an inverse function. And find the inverse function.

Answer to Problem 90E

The domain of function $f$ is $x\ge 1$ for having one-to-one function and inverse function is $f^{-1}\left(x\right)=1-x$.

Explanation of Solution

Given information :

The function $f\left(x\right)=-|x-1|-2$.

Calculation : the domain of function is all real numbers. To having the function one-to-one and inverse, for every value of x , there should be unique value of y.

So the domain of function should be $x\ge 1$.

To find the inverse function substitute $f\left(x\right)=y$ in the function and simplify it,

  $\begin{array}{l}f\left(x\right)=-|x-1|-2\\ y=-|x-1|-2\text{ [substitute }f\left(x\right)=y]\\ -|x-1|=y+2\text{ [add 2 both side]}\\ -x+1=y+2\text{ [remove mod with positive sign because }x\ge 1]\\ -x=y-1\text{ [subtract 1 from both side]}\\ x=1-y\text{ [multiply by }-1]\\ f^{-1}\left(x\right)=1-x\text{ [simplify]}\end{array}$

Thus, the inverse function is $f^{-1}\left(x\right)=1-x$.

The domain of the function f is all real values because function is defined for all real values.

The maximum value of function $-2$ occurs at $x=1$.

Thus, The range of the function f is $(-\infty ,-2]$.

The domain of function $f^{-1}$ is the range of the function f.

Thus, the domain of the function $f^{-1}$ is $(-\infty ,-2]$.

For range of the function $f^{-1}$ , the maximum value $3$ is occur at $x=-2$ ,

Thus, the range of the function $f^{-1}$ is $(-\infty ,3]$.