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