Chapter 1.6, Problem 89E

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 89E

The domain of function $f$ is $x\ge 4$ for having one-to-one function and inverse function is $f^{-1}\left(x\right)=x+3$.

Explanation of Solution

Given information :

The function $f\left(x\right)=|x-4|+1$.

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 $x\ge 4$.

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

  $\begin{array}{l}f\left(x\right)=|x-4|+1\\ y=|x-4|+1\text{ [substitute }f\left(x\right)=y]\\ |x-4|=y-1\text{ [subtract }1\text{ both side]}\\ x-4=y-1\text{ [remove mod with positive sign because }x\ge 4]\\ x=y+3\text{ [add 4 both side and subtract]}\\ f^{-1}\left(x\right)=x+3\text{ [simplify]}\end{array}$

Thus, the inverse function is $f^{-1}\left(x\right)=x+3$.

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

The minimum value of function $1$ occurs at $x=4$.

Thus, The range of the function f is $[1,\infty )$.

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

Thus, the domain of the function $f^{-1}$ is $[1,\infty )$.

For range of the function $f^{-1}$ , the minimum value $4$ is occur at $x=1$ ,

Thus, the range of the function $f^{-1}$ is $[4,\infty )$.