Chapter 1.5, Problem 45E

(a)

To determine

To prove: The function $f$ is its own inverse.

Explanation of Solution

Given information: The function is $f\left(x\right)=\sqrt{1-x^2}$ , $x\ge 0$.

Proof:

If a function is its own inverse then substituting the original function into the function gives value $x$.

Substitute $f\left(x\right)$ for $x$ in the function $f\left(x\right)=\sqrt{1-x^2}$.

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

Simplify further.

  $\begin{array}{c}f\left(f\left(x\right)\right)=\left|x\right|\\ =x\left[\therefore x\le 0\right]\end{array}$

Hence, it is proved that the function $f$ is its own inverse.

(b)

To determine

To prove: The function $f$ is its own inverse.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{1}{x}$.

Proof:

If a function is its own inverse then substituting the original function into the function gives value $x$.

Substitute $f\left(x\right)$ for $x$ in the function $f\left(x\right)=\frac{1}{x}$.

  $\begin{array}{c}f\left(f\left(x\right)\right)=\frac{1}{f\left(x\right)}\\ =\frac{1}{1/x}\\ =x\end{array}$

Hence, it is proved that the function $f$ is its own inverse.