Question

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

Answer 1

Given function $f\left(x\right)=x^3-3x$ is a polynomial function.

$\therefore f\left(x\right)$ is continuous as well as differentiable in $[-\sqrt{3},0].$

Also, $f(-\sqrt{3})={(-\sqrt{3})}^2-3(-\sqrt{3})=-3\sqrt{3}+3\sqrt{3}=0$

And, $f\left(0\right)=0$ Thus, all the three conditions of Rolle's Theorem are satisfied.

$\therefore$ There exist at least one real number c, such that $f^{\prime }\left(c\right)=0$

$f^{\prime }\left(x\right)=3x^2-3$

$\Rightarrow 3c^2-3=0$

$\Rightarrow c^2=1$

$\Rightarrow c=\pm 1$
Thus, $c=-1\in (-\sqrt{3},0)$

Hence, value of $c$ is $-1$.