The value of c in Rolle's theorem when

Solution:

(a) 2

Given:

$f(x)=2 x^{3}-5 x^{2}-4 x+3$

Differentiating the given function with respect to x, we get 

$f^{\prime}(x)=6 x^{2}-10 x-4$

$\Rightarrow f^{\prime}(c)=6 c^{2}-10 c-4$

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

$\Rightarrow 3 c^{2}-5 c-2=0$

$\Rightarrow 3 c^{2}-6 c+c-2=0$

$\Rightarrow 3 c(c-2)+c-2=0$

$\Rightarrow(3 c+1)(c-2)=0$

$\Rightarrow c=2, \frac{-1}{3}$

 

$\therefore c=2 \in\left(\frac{1}{3}, 3\right)$

Thus, $c=2 \in\left(\frac{1}{3}, 3\right)$ for which Rolle's theorem holds.

Hence, the required value of c is 2.