Chapter 4.2, Problem 6E

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 6E

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

Explanation of Solution

Given:

  $f\left(x\right)=\ln \left(x-1\right)$ on $\left[2,4\right]$ .

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,

  $f\left(x\right)=\ln \left(x-1\right)$

It is continuous on $\left[2,4\right]$ and differentiable on $\left(2,4\right)$ .

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

Now,

  $\begin{array}{c}f\left(2\right)=\ln \left(2-1\right)\\ =\ln 1\\ =0\end{array}$

And,

  $\begin{array}{c}f\left(4\right)=\ln \left(4-1\right)\\ =\ln 3\end{array}$

b.

To determine

To find: the value of $c$ .

Answer to Problem 6E

  $c=\frac{2}{\ln 3}+1$ .

Explanation of Solution

Given:

  $f\left(x\right)=\ln \left(x-1\right)$ on $\left[2,4\right]$ .

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 guarantees that there is a point $c$ in the interval $\left(2,4\right)$ for which

  $\begin{array}{c}{f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ \frac{1}{c-1}=\frac{f\left(4\right)-f\left(2\right)}{4-2}\left[\because f^{\prime }\left(x\right)=\frac{1}{x-1}\right]\\ \frac{1}{c-1}=\frac{\ln 3-0}{2}\\ \frac{1}{c-1}=\frac{\ln 3}{2}\\ c-1=\frac{2}{\ln 3}\\ c=\frac{2}{\ln 3}+1\end{array}$