Chapter 4.2, Problem 53E

To determine

To find: the value of the derivative by using the Mean Value Theorem.

Answer to Problem 53E

Option A is correct.

Explanation of Solution

Given:

  $f\left(x\right)=\cos x$ on $\left[0,\frac{\pi }{3}\right]$ .

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,

  $f\left(x\right)=\cos x$

The cosine function is continuous on $\left[0,\frac{\pi }{3}\right]$ and differentiable on $\left(0,\frac{\pi }{3}\right)$ .

Hence the given function satisfies the Hypotheses of the Mean Value Theorem on the given interval.

Now,

  $\begin{array}{c}f\left(0\right)=\cos 0\\ =1\end{array}$

And,

  $\begin{array}{c}f\left(\frac{\pi }{3}\right)=\cos \frac{\pi }{3}\\ =\frac{1}{2}\end{array}$

So, the Mean Value Theorem guarantees that there is a point $c$ in the interval $\left(0,\frac{\pi }{3}\right)$ for which

  $\begin{array}{c}{f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ f^{\prime }\left(c\right)=\frac{f\left(\frac{\pi }{3}\right)-f\left(0\right)}{\frac{\pi }{3}-0}\\ f^{\prime }\left(c\right)=\frac{\frac{1}{2}-1}{\frac{\pi }{3}}\\ f^{\prime }\left(c\right)=\frac{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ f^{\prime }\left(c\right)=-\frac{3}{2\pi }\end{array}$

Hence option A is correct.