Chapter 1.5, Problem 20E

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 20E

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

Calculation:

Substitute $y$ for $f\left(x\right)$ in the given function and solve the equation for $x$ in terms of $y$ for $x\ge -1$ .

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

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 $x\ge -1$ .

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

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

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

Simplify further.

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

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