Chapter 5.2, Problem 8E

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 8E

The function $f\left(x\right)=\{\begin{array}{l}{\sin }^{-1}x-1\le x<1\\ x/2+1\text{ 1}<x\le 3\end{array}$ does not satisfy the hypothesis of Mean Value Theorem.

Explanation of Solution

Given information :

The given function is $f\left(x\right)=\{\begin{array}{l}{\sin }^{-1}x-1\le x<1\\ x/2+1\text{ 1}<x\le 3\end{array}\text{ on }\left[-1,\text{ 3}\right]$

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:

  $f\left(x\right)=\{\begin{array}{l}{\sin }^{-1}x-1\le x<1\\ x/2+1\text{ 1}<x\le 3\end{array}\text{ on }\left[-1,\text{ 3}\right]$

Since, the function $y={\sin }^{-1}x$ and $y=x/2+1$ are continuous function but there might be discontinuity at $x=1$ since the value of the function $f\left(x\right)$ changes from ${\sin }^{-1}x$ to $x/2+1$ .

Also, any function is continuous if $\underset{c\to 1^{-}}{\lim }f\left(x\right)=\underset{c\to 1^{+}}{\lim }f\left(x\right)=f\left(c\right)$ .

Now,

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

Also,

  $\begin{array}{l}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(x/2+1\right)\\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=\left(1/2+1\right)\\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=\frac{3}{2}\end{array}$

Since, $\underset{x\to 1^{-}}{\lim }f\left(x\right)\ne \underset{x\to 1^{+}}{\lim }f\left(x\right)$ , therefore the function is not continuous.

Since, to satisfy the hypothesis of Mean Value Theorem the function $f\left(x\right)$ should be continuous at every point of $\left[-1,\text{ 3}\right]$ . But, here the function $f\left(x\right)$ is not continuous at every point of $\left[-1,\text{ 3}\right]$ .

Hence,

The function $f\left(x\right)=\{\begin{array}{l}{\sin }^{-1}x-1\le x<1\\ x/2+1\text{ 1}<x\le 3\end{array}$ does not satisfy 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 8E

The value of $c$ does not exist.

Explanation of Solution

Given information :

The given function is $f\left(x\right)=x^{1/3}\text{ on }\left[-1,1\right]$

Calculation :

From (a) the function $f\left(x\right)$ does not satisfies the hypothesis of Mean Value Theorem. Therefore, there the value of $c$ does not exist.

Hence,

The value of $c$ does not exist.