Q. 15 - Calculus

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 .

Step-by-Step Solution

Answer

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

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 f x = x 3 - 3 x + 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 are two values of c, and they are both on our interval! 

Step 4. Plot the graph .

The graph of the given function is shown below.