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


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


(a) 1

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

This is a polynomial function, which is continuous and derivable in $R$.

Therefore, the function is continuous on $[0, \sqrt{3}]$ and derivable on $(0, \sqrt{3})$.

Differentiating the given function with respect to x, we get 

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

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

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

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

$\Rightarrow c^{2}=1$


$\Rightarrow c=\pm 1$

Thus, $c=1 \in[0, \sqrt{3}]$ for which Rolle's theorem holds.

Hence, the required value of c is 1.

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