The value of c in Rolle's theorem for the function

Solution:

The given function is $f(x)=x^{3}-3 x$.

f(x) is a polynomial function. We know that a polynomial function is everywhere continuous and differentiable.

So, $f(x)$ is continuous on $[0, \sqrt{3}]$ and differentiable on $(0, \sqrt{3})$.

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

$\therefore f(0)=f(\sqrt{3})$

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

So, there exist a real number $c \in(0, \sqrt{3})$ such that $f^{\prime}(c)=0$.

$f(x)=x^{3}-3 x$

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

$\therefore f^{\prime}(c)=0$

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

$\Rightarrow c^{2}=1$

 

$\Rightarrow c=\pm 1$

Thus, $c=1 \in(0, \sqrt{3})$ such that $f^{\prime}(c)=0$.

Hence, the value of c is 1.

The value of $c$ in Rolle's theorem for the function $f(x)=x^{3}-3 x$ in the interval $[0, \sqrt{3}]$ is ___1___.