Question

A function f defined on [−3,−1] with f (−3) = f (−1) = 0 such that f is continuous everywhere except at x = −1 and differentiable everywhere except at x = −1, and fails the conclusion of Rolle’s Theorem.

Short answer

It is not satisfied the condition of Roll's theorem on$\left[-3,-1\right]$.

Step-by-step solution

Step 1. Given information .

Consider the given function f defined on [−3,−1] with f (−3) = f (−1) = 0

such that f is continuous everywhere except at x = −1 and differentiable everywhere except at x = −1, and fails the conclusion of Rolle’s Theorem.

Step 2. Using Roll'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) forwhich 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 (a) = f (b) = -3, or if f (a) = f (b) = −1, 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$f\text{'}\left(c\right)=0$.