Chapter 0.5, Problem 45E

(a.)

To determine

To Prove: The given function $f$ is its own inverse.

Answer to Problem 45E

It has been shown that the given function $f$ is its own inverse.

Explanation of Solution

Given:

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

Concept used:

The inverse of a function $f\left(x\right)$ can be obtained by assuming $y=f\left(x\right)$ and then solving for $x$ in terms of $y$ .

Calculation:

The given function is,

  $f\left(x\right)=\sqrt{1-x^2}$ , $x\ge 0$

Let $y=f\left(x\right)$ . Put this value in the above expression to get,

  $y=\sqrt{1-x^2}$

Squaring both sides,

  $y^2=1-x^2$

Simplifying,

  $x^2=1-y^2$

Taking square root on both sides,

  $x=\sqrt{1-y^2}$

Now, $y=\sqrt{1-x^2}$ , which is the positive square root.

This implies that $y\ge 0$ .

As assumed previously,

  $y=f\left(x\right)$

Then,

  $x=f^{-1}\left(y\right)$

Put this value in $x=\sqrt{1-y^2}$ to get,

  $f^{-1}\left(y\right)=\sqrt{1-y^2}$

So, it follows that,

  $f^{-1}\left(y\right)=\sqrt{1-y^2}$ , $y\ge 0$

Replacing the variable $y$ by $x$ ,

  $f^{-1}\left(x\right)=\sqrt{1-x^2}$ , $x\ge 0$

Comparing with the given function; $f\left(x\right)=\sqrt{1-x^2}$ , $x\ge 0$ ; it follows that,

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

This shows that the given function $f$ is its own inverse.

Conclusion:

It has been shown that the given function $f$ is its own inverse.

(b.)

To determine

To Prove: The given function $f$ is its own inverse.

Answer to Problem 45E

It has been shown that the given function $f$ is its own inverse.

Explanation of Solution

Given:

The function, $f\left(x\right)=\frac{1}{x}$ .

Concept used:

The inverse of a function $f\left(x\right)$ can be obtained by assuming $y=f\left(x\right)$ and then solving for $x$ in terms of $y$ .

Calculation:

The given function is,

  $f\left(x\right)=\frac{1}{x}$

Let $y=f\left(x\right)$ . Put this value in the above expression to get,

  $y=\frac{1}{x}$

Simplifying,

  $x=\frac{1}{y}$

As assumed previously,

  $y=f\left(x\right)$

Then,

  $x=f^{-1}\left(y\right)$

Put this value in $x=\frac{1}{y}$ to get,

  $f^{-1}\left(y\right)=\frac{1}{y}$

Replacing the variable $y$ by $x$ ,

  $f^{-1}\left(x\right)=\frac{1}{x}$

Comparing with the given function; $f\left(x\right)=\frac{1}{x}$ ; it follows that,

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

This shows that the given function $f$ is its own inverse.

Conclusion:

It has been shown that the given function $f$ is its own inverse.