Question

A functionf that is defined on [−2, 2] with f (−2) = f (2) = 0 such that f is continuous everywhere, differentiable everywhere except at x = −1, and fails the conclusion of Rolle’s Theorem .

Short answer

It is not satisfied the conditions of Rolle's theorem on$\left[-2,2\right]$.

Step-by-step solution

Step 1. Given information .

Consider the given function that f is continuous everywhere, differentiable everywhere except at x = −1, and fails the conclusion of

Rolle’s Theorem.

Step 2. Using the Rolle's theorem . 

If f is continuous on [a, b] and differentiable on (a, b), and if f (a) = f (b) = 0, then there exists at least one value c ∈ (a, b) for which f '(c) = 0 .

Step 3. Classify the Rolle's theorem . 

Actually, Rolle’s Theorem also holds in the more general case where f (a) and f (b) are equal to each other (not necessarily both zero). For example, Rolle’s Theorem is also true if f (-2) = f (2) = 0, and so on, because vertically shifting a function by adding a constant term does not change its derivative. However, the classic way to state Rolle’s Theorem is with f (a) and f (b) both equal to zero. If f (a) is not equal to f (b) then in this condition the conclusion of Rolle's theorem fails . There are three conditions that will be satisfied .

(a) f is continuous on $\left[a,b\right]$.

(b) f is differentiable on $\left(a,b\right)$.

(c) f (a) = f (b)

then there exist at least one point c in open interval $\left(a,b\right)$that is$f\text{'}\left(c\right)=0$.