Chapter 6.7, Problem 35PPSE

Solution

To find: The inverse of the function $f\left(x\right)=-x^2+1$ , domain and range of the function and its inverse and determine if the inverse is a function or not.

The inverse of the function is $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ and it’s not a function.

Domain of $f\left(x\right)=-x^2+1$ : $\left(-\infty ,\infty \right)$

Range of $f\left(x\right)=-x^2+1$ : $\left(\infty ,1\right]$

Domain of $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ : $\left(\infty ,1\right]$

Range of $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ : $\left(-\infty ,\infty \right)$

Given information:

The function, $f\left(x\right)=-x^2+1$ .

Concept used:

Domain of inverse of a function is the range of the function and the range of the inverse is the domain of the function.

Inverse $f^{-1}\left(x\right)$ of a function $f\left(x\right)$ is a function if for every $x$ in the domain of $f^{-1}\left(x\right)$ there is only one $y$ value in the range of $f^{-1}\left(x\right)$ .

Calculation:

Find the inverse of the function $f\left(x\right)=-x^2+1$ .

  $\begin{array}{c}f\left(x\right)=-x^2+1\\ y=-x^2+1\text{Rewrite the function with }y\\ x=-y^2+1\text{Switch }x\text{ and }y\\ -y^2+1=x\text{Solve for }y\\ -y^2=x-1\\ y^2=-x+1\\ y=\pm \sqrt{-x+1}\end{array}$

So, $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ .

So, the domain of the function $f\left(x\right)=-x^2+1$ is $\left(-\infty ,\infty \right)$ .

Since $x^2\ge 0$ , $-x^2+1\le 1$ . So, the range of the function $f\left(x\right)=-x^2+1$ is $\left(\infty ,1\right]$ .

Since the domain of inverse of a function is the range of the function and the range of the inverse is the domain of the function, the domain of the inverse, $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ , will be $\left(\infty ,1\right]$ and the range will be $\left(-\infty ,\infty \right)$ .

The inverse is not a function since for every $x$ in the domain of $f^{-1}\left(x\right)$ , there are two $y$ values in the range of $f^{-1}\left(x\right)$ .

Conclusion:

The inverse of the function is $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ and it’s not a function.

Domain of $f\left(x\right)=-x^2+1$ : $\left(-\infty ,\infty \right)$

Range of $f\left(x\right)=-x^2+1$ : $\left(\infty ,1\right]$

Domain of $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ : $\left(\infty ,1\right]$

Range of $f^{-1}\left(x\right)=\pm \sqrt{-x+1}$ : $\left(-\infty ,\infty \right)$