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

- f is continuous on the close interval [a, b]
- f is differentiable on the open interval (a, b)
- f(a) = f(b)

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

**Problem Part:**

Solution:

Given

f(x) = x^{3} – x^{2} – 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]

- f(x) is continuous on the close interval [0, 3]
- f
^{‘}(x) = 3x^{2}– 2x – 6 exists in (0, 3). So, f(x) is differentiable in (0, 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, 3c^{2} – 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.

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