Chapter 5.2, Problem 60E

To determine

To show : For $f\left(x\right)=x^2$ in the interval $\left[a,b\right]$ the conclusion of the Mean Value Theorem is $c=\frac{a+b}{2}$ .

Answer to Problem 60E

The possible explanation is shown below.

Explanation of Solution

Given information :

Two $a$ and $b$ and the function $f\left(x\right)=x^2$ in the interval $\left[a,b\right]$ .

Calculation :

The Mean Value Theorem states that if $f\left(x\right)$ 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 exists at least one $c$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .

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

Since, $f\left(x\right)=x^2$ , so $f\left(x\right)$ is continuous and differentiable for all $x$ . Therefore, by Mean Value Theorem there must be at least $c$ such that

  $\begin{array}{l}f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ f\text{'}\left(c\right)=\frac{b^2-a^2}{b-a}\\ f\text{'}\left(c\right)=\frac{\left(b-a\right)\left(b+a\right)}{b-a}\\ f\text{'}\left(c\right)=a+b\mathrm{......}\text{ (1)}\end{array}$

Differentiate $f\left(x\right)=x^2$ with respect to $x$ and replace $x$ with $c$ :

  $\begin{array}{l}f\left(x\right)=x^2\\ f\text{'}\left(x\right)=2x\\ f\text{'}\left(c\right)=2c\end{array}$

Substitute $f\text{'}\left(c\right)$ in equation (1) and solve for $c$ :

  $\begin{array}{l}2c=a+b\\ c=\frac{a+b}{2}\end{array}$