Chapter 4.2, Problem 57E

To determine

To find: the value of $c$ in the conclusion of the Mean Value Theorem.

Answer to Problem 57E

  $c=\sqrt{ab}$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{1}{x}$ on $\left[a,b\right]$ , where $a$ and $b$ are positive numbers.

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,

  $f\left(x\right)=\frac{1}{x}$

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

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

Now,

  $f\left(a\right)=\frac{1}{a}$

And,

  $f\left(b\right)=\frac{1}{b}$

Now, the Mean Value Theorem guarantees that there is a point $c$ in the interval $\left(a,b\right)$ for which

  $\begin{array}{c}{f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ -\frac{1}{c^2}=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}\left[\because f^{\prime }\left(x\right)=-\frac{1}{x^2}\right]\\ -\frac{1}{c^2}=\frac{\frac{a-b}{ab}}{b-a}\\ -\frac{1}{c^2}=\frac{a-b}{ab\left(b-a\right)}\\ -\frac{1}{c^2}=\frac{a-b}{-ab\left(a-b\right)}\\ \frac{1}{c^2}=\frac{1}{ab}\\ c^2=ab\\ c=\sqrt{ab}\end{array}$