Question

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

Function $f\left(x\right)=x^3-3x+2$satisfied all the conditions of Rolle's theorem .

Step-by-step solution

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.

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.

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 $\left(a,b\right)$that is $f\text{'}\left(c\right)=0$.

Differentiate the function.

$$\begin{gathered} f\left(x\right)=x^3-3x+2 \\ f\text{'}\left(x\right)=3x^2-3 \end{gathered}$$

Put $f\text{'}\left(c\right)=0$.

$$\begin{gathered} 3c^2-3=0 \\ 3c^2=3 \\ c=\pm 1 \end{gathered}$$

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

Step 4. Plot the graph .

The graph of the given function is shown below.