Question

A function f defined on [−2, 0] such that f is continuous everywhere except at x = −2, differentiable everywhere except at x = −2, and fails the conclusion of the Mean Value Theorem with a = −2 and b = 0.

Short answer

The function is continuous and differentiable at $x=-2$but did not satisfy the mean value theorem .

Step-by-step solution

Step 1. Given information .

Consider the given function f is continuous everywhere except at x = −2, differentiable everywhere except at x = −2, and fails the conclusion of the Mean value theorem .

Step 2. Using 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 $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

$$\begin{gathered} f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a} \\ f\text{'}\left(c\right)=\frac{0-\left(-2\right)}{0-\left(-2\right)} \\ f\text{'}\left(c\right)=\frac{2}{2}=1 \end{gathered}$$

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 .