The value of c in Rolle's theorem when $f(x)=2 x^3-5 x^2-4 x+3, x \in[1 / 2,3]$ is

2

Clearly, $f(x)$ being a polynomial is continuous on $[1 / 2,3]$ and differentiable on $(1 / 2 , 3)$.

Also, $f\left(\frac{1}{2}\right)=f(3)=0$

So, by Rolle's theorem there exists $c \in(1 / 2,3)$ such that

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

$\Rightarrow 6 c^2-10 c-4=0$         $\left[∵ f^{\prime}(x)=6 x^2-10 x-4\right]$

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

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