Question

Find the value of c in Rolle's Theorem for the function f(x)=$x^3-3xin[-\sqrt{3},0].$

Answer 1

Since the polynomial function f(x)=$x^3-3x$ is everywhere continuous and differentiable,so

i) f(x) is continuous on $[-\sqrt{3},0]$ and,

ii) f(x) is differentiable on $(-\sqrt{3},0)$.

iii) Also $f(-\sqrt{3})=-3\sqrt{3}+3\sqrt{3}=0$ and f(0)=0 $\Rightarrow f(-\sqrt{3})=f\left(0\right)$

$\therefore$ all the conditions of Rolle's theorem are satisfied.So there must exist one point $cϵ(-\sqrt{3},0)$ such that f'(c)=0.Now f'(x)=$3x^2-3$
For f'(c)=0,$3c^2-3=0$ $\therefore c=-1ϵ(-\sqrt{3},0)$