Chapter 1.5, Problem 16E

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 16E

The inverse of the given function $f^{-1}\left(x\right)=\sqrt{x-1}$ 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 $f\left(x\right)=x^2+1$ , $x\ge 0$ .

Calculation:

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

  $\begin{array}{c}y=x^2+1\\ x^2=y-1\\ x=\sqrt{y-1}\end{array}$

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

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

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 $\sqrt{x-1}$ for $f^{-1}\left(x\right)$ to find $\left(f\circ f^{-1}\right)\left(x\right)$ for $x\ge 0$ .

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

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 $x^2+1$ for $f\left(x\right)$ in to find $\left(f^{-1}\circ f\right)\left(x\right)$ for $x\ge 0$ .

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

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