Chapter 4.2, Problem 58E

To determine

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

Answer to Problem 58E

  $c=\frac{a+b}{2}$ .

Explanation of Solution

Given:

  $f\left(x\right)=x^2$ on $\left[a,b\right]$ .

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,

  $f\left(x\right)=x^2$

Since it is a polynomial function, therefore 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,

  $f\left(a\right)=a^2$

And,

  $f\left(b\right)=b^2$

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

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