A function f that satisfies the hypotheses of Rolle’s Theorem on [−2, 2] and for which there are exactly three values c ∈ (−2, 2) that satisfy the conclusion of the theorem .

Short Answer

Expert verified

Function fx=x3-3x+2satisfied all the conditions of Rolle's theorem .

Step by step solution

01

Step 1. Given information .

Consider the function f that satisfies the hypotheses of Rolle’s Theorem on [−2, 2] and for which there are exactly three values c ∈ (−2, 2) that satisfy the conclusion of the theorem.

02

Step 2. Using Rolle's theorem .

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.

03

Step 3. Classify Rolle's theorem for function fx=x3-3x+2.

The main conditions of the Mean Value Theorem are:

(a) f(x) must be continuous on [a, b]

(b) f(x) must be differentiable on (a, b).

(c) There is some point c on a,bthat is f'c=0.

Differentiate the function.

fx=x3-3x+2f'x=3x2-3

Put f'c=0.

3c2-3=03c2=3c=±1

As it turns out, there aretwo values of c, and they areboth on our interval!

04

Step 4. Plot the graph .

The graph of the given function is shown below.

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