Find the value of c in Rolle’s theorem for the function f(x) = x³ – 3x in

We have-

f(x) = x³-3x

Since a polynomial function is everywhere continuous and

differentiable, therefore f(x) is continuous on [-√3, 0] and

differentiable on [-√3, 0].

Also, f(-√3) = (-√3)³ - 3(-√3) = -3√3 + 3√3 = 0

and, f(0) = (0)³ - 3(0) = 0

∴ f(-√3) = f(0)

Thus, all the three conditions of Rolle's theorem are satisfied.

Thus, there exists c ∈ (-√3, 0) such that f'(c) = 0

Now, f'(x) = 3x² - 3

∴ f'(x) = 0

⇒ 3x² - 3 = 0

⇒ 3(x² - 1) = 0

⇒ x² - 1 = 0

∴ x = ± 1

Clearly, only -1 lies in the interval (-√3, 0).

Thus, the value of c is -1.