# State Rolle’s theorem and verify the Rolle’s theorem for f(x) = x3 – x2 – 6x + 2 in [0, 3]

Definition: Let f be a function that satisfies the following three hypothesis:

1. f is continuous on the close interval [a, b]
2. f is differentiable on the open interval (a, b)
3. f(a) = f(b)

Then there is a number c in (a, b) such that f(c) = 0

Problem Part:

Solution:

Given

f(x) = x3 – x2 – 6x + 2 on [0, 3]

since, $$\lim_{x \to 0}f(x) = 2$$ and $$\lim_{x \to 3^{-}}f(x) = 2$$

Thus, f(x) is continuous at end points and all interior points of [0, 3]

1. f(x) is continuous on the close interval [0, 3]
2. f(x) = 3x2 – 2x – 6 exists in (0, 3). So, f(x) is differentiable in (0, 3).
3. f(0) = f(3) = 2

Thus, all three conditions are holds on f(x). Hence, there exists  point $$c \in (0, 3)$$ such that

f(c) = 0

or, 3c2 – 2c – 6 = 0

$$or, c = \frac{1 \pm \sqrt{19}}{3}$$

$$or, c = \frac{1 + \sqrt{19}}{3} \in (0, 3)$$

So, Rolle’s theorem is verified.