Chapter 4, Problem 525RE

True or False? Justify your answer with a proof or a counterexample. Assume that f(x) is continuous and differentiable unless stated otherwise.

525. If $f\left(-1\right)=-6$ and $f\left(1\right)=2$ , then there exists at least one point $x\in \left[-1,1\right]$ such that $f\text{'}\left(x\right)=4$ .

To determine

To check: whether the given data is true or false if $f\left(x\right)$ is continuous and differentiable.

Answer to Problem 525RE

The statement is true.

Explanation of Solution

Given information:

The given data is $f\left(-1\right)=-6\text{and}f\left(1\right)=2$ .

Then, there exists at least one point $x\in \left[1,1\right]$ such that $f^{\prime }\left(x\right)=4$ .

Formula used:

Mean Value Theorem:

If $f\left(x\right)$ is continuous in closed interval $\left[a,b\right]$ and differentiable in open interval $\left(a,b\right)$ ,

then there exists some $x\in \left[a,b\right]$ , such that

  $f^{\prime }\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

Calculation:

The given data is $f\left(-1\right)=-6\text{and}f\left(1\right)=2$ .

Then, there exists at least one point $x\in \left[1,1\right]$ such that $f^{\prime }\left(x\right)=4$ .

Here, $f\left(a\right)=f\left(-1\right)=-6\text{and}f\left(b\right)=f\left(1\right)=2$ .

Thus, by the mean value theorem,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ =\frac{2-\left(-6\right)}{1-\left(-1\right)}\\ =\frac{8}{2}\\ =4\end{array}$

Therefore, by the mean value theorem the given statement is correct.

Conclusion:

Hence, the given statement is true.