Chapter 1.6, Problem 87E

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 87E

The domain of function $f$ is $x\ge 0$ for having one-to-one function and inverse function is $f^{-1}\left(x\right)=\sqrt{-\frac{x+5}{2}}$.

Explanation of Solution

Given information :

The function $f\left(x\right)=-2x^2-5$.

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 $x\ge 0$.

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

  $\begin{array}{l}f\left(x\right)=-2x^2-5\\ y=-2x^2-5\text{ [substitute }f\left(x\right)=y]\\ -2x^2=y+5\text{ [add 5 both side]}\\ x^2=\frac{y+5}{-2}\text{ [divide by }-2\text{]}\\ x=\sqrt{-\frac{y+5}{2}}\text{ [sqaure root]}\\ f^{-1}\left(x\right)=\sqrt{-\frac{x+5}{2}}\text{ [simplify]}\end{array}$

Thus, the inverse function is $f^{-1}\left(x\right)=\sqrt{-\frac{x+5}{2}}$.

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

The maximum value of function $-5$ occurs at $x=0$.

Thus, The range of the function f is $(-\infty ,-5]$.

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

Thus, the domain of the function $f^{-1}$ is $(-\infty ,-5]$.

For range of the function $f^{-1}$ , the maximum value $0$ is occur at $x=-5$ ,

Thus, the range of the function $f^{-1}$ is $(-\infty ,0]$.