Chapter 1.6, Problem 81E

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 81E

The domain of function $f$ is $x\ge 2$ for having one-to-one function and inverse function is $f^{-1}\left(x\right)=\sqrt{x}+2$.

Explanation of Solution

Given information :

The function $f\left(x\right)={\left(x-2\right)}^2$.

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 $x\ge 2$.

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

  $\begin{array}{l}f\left(x\right)={\left(x-2\right)}^2\\ y={\left(x-2\right)}^2\text{ [substitute }f\left(x\right)=y]\\ x-2=\sqrt{y}\text{ [square root]}\\ x=\sqrt{y}+2\text{ [add 2 both side]}\\ f^{-1}\left(x\right)=\sqrt{x}+2\text{ [simplify]}\end{array}$

Thus, the inverse function is $f^{-1}\left(x\right)=\sqrt{x}+2$.

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

The range of the function f is $[0,\infty )$.

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

Thus, the domain of the function $f^{-1}$ is $[0,\infty )$.

For range of the function $f^{-1}$ , the minimum value $2$ is occur at $x=0$ ,

Thus, the range of the function $f^{-1}$ is $[2,\infty )$.