Chapter 3.3, Problem 38E

(a)

To determine

To find: the domain of $f$ .

Answer to Problem 38E

The domain of function $f$ is $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information :

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

  

On observing the graph of function, the domain of the function is whole real line.

The domain of function $f$ is $\left(-\infty ,\infty \right)$ .

(b)

To determine

To find: the range of $f$ .

Answer to Problem 38E

The range of function $f$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ .

Explanation of Solution

Given information :

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

  

On observing the graph of function, the range of the function is lies in between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$ .

The range of function $f$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ .

(c)

To determine

To find: the points where $f$ is not differentiable.

Answer to Problem 38E

The function $f$ is not differentiable at $x=\frac{\left(2n-1\right)\pi }{2},n\in \mathbb{Z}$ .

Explanation of Solution

Given information :

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

  

On observing the graph of function, there are certain points like $\frac{\pi }{2},\frac{3\pi }{2},\mathrm{...}$ where the graph of function has different values when the left and right limit approaches to that point.

In general, the points are $x=\frac{\left(2n-1\right)\pi }{2},n\in \mathbb{Z}$

(d)

To determine

To sketch: the graph of derivative of $f$ .

Explanation of Solution

Given information :

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

  

The derivative graph of $f$ is

  

(e)

To determine

To find: derivative of $f$ algebraically.

Answer to Problem 38E

  $f^{\prime }\left(x\right)=\frac{\cos x}{\sqrt{1-{\sin }^{2}x}}$

Explanation of Solution

Given information :

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

  

Differentiate $f\left(x\right)={\sin }^{-1}\left(\sin x\right)$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}{\sin }^{-1}\left(\sin x\right)\\ =\frac{1}{\sqrt{1-{\left(\sin x\right)}^2}}\cdot \frac{d}{dx}\left(\sin x\right)\left[\frac{d}{dx}{\sin }^{-1}\left(x\right)=\frac{1}{\sqrt{1-x^2}}\right]\\ =\frac{1}{\sqrt{1-{\sin }^{2}x}}\cdot \left(\cos x\right)\\ =\frac{\cos x}{\sqrt{1-{\sin }^{2}x}}\end{array}$

Hence, $f^{\prime }\left(x\right)=\frac{\cos x}{\sqrt{1-{\sin }^{2}x}}$ .

The derivative graph and derivative algebraically both are same.