Chapter 5.2, Problem 1E

a.

To determine

To state : Whether or not the function satisfies the hypothesis of the Mean Value Theorem on the given interval.

Answer to Problem 1E

The function $f\left(x\right)=x^2+2x-1$ satisfies the hypothesis of Mean Value Theorem.

Explanation of Solution

Given information :

The given function is $f\left(x\right)=x^2+2x-1\text{ on }\left[0,1\right]$

Calculation :

The function $f\left(x\right)$ satisfies the hypothesis of the Mean Value Theorem 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 $\left(a,b\right)$ .

Consider the given function:

  $f\left(x\right)=x^2+2x-1\text{ on }\left[0,1\right]$

Since, the above function $f\left(x\right)$ is polynomial function so it is continuous on $\left[0,1\right]$ and differentiable on $\left(0,\text{ 1}\right)$ .

Therefore, the function $f\left(x\right)$ satisfies the hypothesis of Mean Value Theorem.

Hence,

The function $f\left(x\right)=x^2+2x-1$ satisfies the hypothesis of Mean Value Theorem.

b.

To determine

To find : Value of $c$ in the interval $\left(a,b\right)$ that satisfies the equation $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .

Answer to Problem 1E

The value of $c$ is $\frac{1}{2}$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)=x^2+2x-1\text{ on }\left[0,1\right]$

Calculation :

From (a) the function $f\left(x\right)$ satisfies the satisfies the hypothesis of Mean Value Theorem. Therefore, the Mean Value Theorem guarantees a point $c$ in the interval $\left(0,\text{ 1}\right)$ for which

  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, $a=0$ and $b=1$ , so

  $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$ ...... (1)

Now,

  $\begin{array}{l}f\left(1\right)={\left(1\right)}^2+2\left(1\right)-1\\ f\left(1\right)=1+2-1\\ f\left(1\right)=2\end{array}$

And

  $\begin{array}{l}f\left(0\right)={\left(0\right)}^2+2\left(0\right)-1\\ f\left(0\right)=0+0-1\\ f\left(0\right)=-1\end{array}$

Now, find the derivative of the function $f\left(x\right)$ and replace the $x$ with $c$ to get $f\text{'}\left(c\right)$ :

  $\begin{array}{l}f\left(x\right)=x^2+2x-1\\ f\text{'}\left(x\right)=2x+2\\ f\text{'}\left(c\right)=2c+2\end{array}$

Substitute the values of $f\left(1\right)$ , $f\left(0\right)$ and $f\text{'}\left(c\right)$ in equation (1) and solve for $c$ :

  $\begin{array}{l}2c+2=\frac{2-\left(-1\right)}{1-0}\\ 2c+2=3\\ 2c=3-2\\ c=\frac{1}{2}\end{array}$

Hence,

The value of $c$ is $\frac{1}{2}$ .