Question:
The value of c in Rolle's theorem when
$f(x)=2 x^{3}-5 x^{2}-4 x+3, x \in[1 / 3,3]$ is
(a) 2
(b) $-\frac{1}{3}$
(c) $-2$
(d) $\frac{2}{3}$
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.