Question

A function f defined on [−3, 3] such that f is continuous everywhere except at x = 1, differentiable everywhere except at x = 1, and fails the conclusion of the Mean Value Theorem with a = −3 and b = 3.

Short answer

The function is continuous and differentiable at $\left[-3,3\right]$but did not satisfy the mean value theorem .

Step-by-step solution

Step 1. Given information .

Consider the function f is continuous everywhere except at x = 1, differentiable everywhere except at x = 1, and fails the conclusion of the Mean Value theorem .

Step 2. Using  the Mean value theorem .

If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one value c ∈ (a, b) such that$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Step 3. Classify the Mean value theorem . 

To classify the mean value theorem substitute the value of a and b in the given theorem .

$$\begin{gathered} f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a} \\ f\text{'}\left(c\right)=\frac{-3-3}{-3-3} \\ f\text{'}\left(c\right)=\frac{-6}{-6} \\ f\text{'}\left(c\right)=-1 \end{gathered}$$

Therefore the value of $f\text{'}\left(c\right)$is not equal to zero the given statement does not satisfy the conditions of Mean value theorem .