Chapter 4.2, Problem 7E

a.

To determine

To state: whether the given function satisfies the hypotheses of the Mean Value Theorem or not in the given interval.

Answer to Problem 7E

The given function does not satisfy the Hypotheses of the Mean Value Theorem on the given interval.

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{l}\cos x,0\le x<\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \sin x,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\le x\le \pi \end{array}\right.$ on $\left[0,\pi \right]$ .

Concept used:

Mean Value Theorem for Derivatives:

If $y=f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there is at least one point $c$ in $\left(a,b\right)$ at which the instantaneous rate of change equals the mean rate of change,

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Consider the given function,

  $f\left(x\right)=\left\{\begin{array}{l}\cos x,0\le x<\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \sin x,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\le x\le \pi \end{array}\right.$

The given function is not continuous on $\left[0,\pi \right]$ , because it is not continuous at $x=\frac{\pi }{2}$ . Because,

  $\begin{array}{c}\underset{x\to {\frac{\pi }{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{-}}{\lim }\cos x\\ =\cos \frac{\pi }{2}\\ =0\end{array}$

And,

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

Hence the given function does not satisfy the Hypotheses of the Mean Value Theorem on the given interval.

b.

To determine

To find: the value of $c$ .

Answer to Problem 7E

There is no value of $c$ .

Explanation of Solution

Given:

  $f\left(x\right)=\left\{\begin{array}{l}\cos x,0\le x<\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \sin x,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\le x\le \pi \end{array}\right.$ on $\left[0,\pi \right]$ .

Concept used:

Mean Value Theorem for Derivatives:

If $y=f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there is at least one point $c$ in $\left(a,b\right)$ at which the instantaneous rate of change equals the mean rate of change,

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

The Mean Value Theorem do not guarantees that there is a point $c$ in the interval $\left(0,4\right)$ for which

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .