Chapter 1.5, Problem 19E

To determine

To find: The inverse of the given function and also verify that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Answer to Problem 19E

The inverse of the given function is $f^{-1}\left(x\right)=2-\sqrt{-x}$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=-{\left(x-2\right)}^2$ , $x\le 2$ .

Calculation:

Substitute $y$ for $f\left(x\right)$ in the given function and solve the equation for $x$ in terms of $y$ for $x\le 2$ .

  $\begin{array}{c}y=-{\left(x-2\right)}^2\\ {\left(x-2\right)}^2=-y\\ x-2=-\sqrt{-y}\\ x=2-\sqrt{-y}\end{array}$

Interchange $x$ and $y$ , and substitute $f^{-1}\left(x\right)$ for $y$ .

  $\begin{array}{c}y=2-\sqrt{-x}\\ f^{-1}\left(x\right)=2-\sqrt{-x}\end{array}$

Verify:

The composite function $\left(f\circ f^{-1}\right)\left(x\right)$ can be written as:

  $\left(f\circ f^{-1}\right)\left(x\right)=f\left(f^{-1}\left(x\right)\right)$

Substitute $2-\sqrt{-x}$ for $f^{-1}\left(x\right)$ to find $\left(f\circ f^{-1}\right)\left(x\right)$ for $x\le 2$ .

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=f\left(2-\sqrt{-x}\right)\\ =-{\left(2-\sqrt{-x}-2\right)}^2\\ =-{\left(-\sqrt{-x}\right)}^2\\ =x\end{array}$

The composite function $\left(f^{-1}\circ f\right)\left(x\right)$ can be written as:

  $\left(f^{-1}\circ f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)$

Substitute $-{\left(x-2\right)}^2$ for $f\left(x\right)$ to find $\left(f^{-1}\circ f\right)\left(x\right)$ for $x\le 2$ .

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)=f^{-1}\left(-{\left(x-2\right)}^2\right)\\ =2-\sqrt{{\left(x-2\right)}^2}\\ =x\end{array}$

It is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Therefore, the inverse of the given function is $f^{-1}\left(x\right)=2-\sqrt{-x}$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .