Chapter 4.2, Problem 1E

To determine

To estimate the numbers in $\left[1.10\right]$ that satisfy the conclusion of the mean value theorem.

Answer to Problem 1E

3, 7 are the numbers in $\left[1.10\right]$ that satisfy the conclusion of the mean value theorem.

Explanation of Solution

Given information:

Graph $f$

  

Concept Used:

Mean Value Theorem:

If f is continuous on a closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$ , then there exists a number $c$ in $\left(a,b\right)$ such that

  $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

Calculation:

We have to find the numbers in $\left[1,10\right]$ that satisfy the mean value theorem·

If we take two ordered pair $\left(a,f\left(a\right)\right)$ and $\left(a,f\left(b\right)\right)$ ,

Then slope = $\frac{f(b)-f(a)}{b-a}$

Which is equal to $f^{\prime }(c)$ as per mean value theorem·

So, $f^{\prime }(c)$ is nothing but a slope·

Let’s refer the figure below,

  

  $(a,f(a))=(1,f(1))$ , $(b,f(b))=(10,f(10))$ · So, the slope of Y is equal to the line M and N since both are parallel to Y· Look at the figure· The line M intersect the line Y at 3 and line N intersect the line Y at 7 approximately·

Slope of $Y=\frac{f\left(10\right)-f\left(1\right)}{10-1}=f^{\prime }\left(c\right)$

Since the slope of M and N are equal to Y,

So,$\begin{array}{l}{f}^{\prime }\left(3\right)=\frac{f\left(10\right)-f\left(1\right)}{10-1}\\ f^{\prime }\left(7\right)=\frac{f\left(10\right)-f\left(1\right)}{10-1}\end{array}$

Then, $c$ is 3 and 7.

Thus,

3 and 7 are the numbers in $\left[1,10\right]$ that satisfy the mean value theorem·