Chapter 5.2, Problem 5E

a.

To determine

To state : Whether or not the function satisfies the hypothesis of the Mean Value Theorem on the given interval.

Answer to Problem 5E

The function $f\left(x\right)={\sin }^{-1}x$ satisfies the hypothesis of Mean Value Theorem.

Explanation of Solution

Given information :

The given function is $f\left(x\right)={\sin }^{-1}x\text{ on }\left[-1,1\right]$

Calculation :

The function $f\left(x\right)$ satisfies the hypothesis of the Mean Value Theorem if $f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ .

Consider the given function:

  $f\left(x\right)={\sin }^{-1}x\text{ on }\left[-1,1\right]$

Since, the above function $f\left(x\right)$ is inverse trigonometric function so it is continuous on $\left[-1,1\right]$ .

Differentiate $f\left(x\right)$ with respect to $x$ :

  $\begin{array}{l}f\left(x\right)={\sin }^{-1}x\\ f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}}\end{array}$

Therefore, the function is differentiable on $\left(-1,\text{ 1}\right)$ .

Hence,

The function $f\left(x\right)={\sin }^{-1}x$ satisfies the hypothesis of Mean Value Theorem.

b.

To determine

To find : Value of $c$ in the interval $\left(a,b\right)$ that satisfies the equation $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .

Answer to Problem 5E

The value of $c$ is $\sqrt{1-\frac{4}{{\pi }^2}}$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)={\sin }^{-1}x\text{ on }\left[-1,1\right]$

Calculation :

From (a) the function $f\left(x\right)$ satisfies the hypothesis of Mean Value Theorem. Therefore, the Mean Value Theorem guarantees a point $c$ in the interval $\left(-1,\text{ 1}\right)$ for which

  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, $a=-1$ and $b=1$ , so

  $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(-1\right)}{1-\left(-1\right)}$ ...... (1)

Now,

  $\begin{array}{l}f\left(-1\right)={\sin }^{-1}\left(-1\right)\\ f\left(-1\right)=-\frac{\pi }{2}\end{array}$

And

  $\begin{array}{l}f\left(1\right)={\sin }^{-1}\left(1\right)\\ f\left(1\right)=\frac{\pi }{2}\end{array}$

Now, find the derivative of the function $f\left(x\right)$ and replace the $x$ with $c$ to get $f\text{'}\left(c\right)$ :

  $\begin{array}{l}f\left(x\right)={\sin }^{-1}x\\ f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}}\\ f\text{'}\left(c\right)=\frac{1}{\sqrt{1-c^2}}\end{array}$

Substitute the values of $f\left(-1\right)$ , $f\left(1\right)$ and $f\text{'}\left(c\right)$ in equation (1) and solve for $c$ :

  $\begin{array}{l}\frac{1}{\sqrt{1-c^2}}=\frac{\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)}{1-\left(-1\right)}\\ \sqrt{1-c^2}=\frac{2}{\pi }\\ 1-c^2=\frac{4}{{\pi }^2}\\ c^2=1-\frac{4}{{\pi }^2}\\ c=\sqrt{1-\frac{4}{{\pi }^2}}\end{array}$

Hence,

The value of $c$ is $\sqrt{1-\frac{4}{{\pi }^2}}$ .