$x \in[-4,2]$ માં વિધેય $f(x)=x^{2}+2 x-8$ માટે રોલનું પ્રમેય ચકાસો. 

The given function, $f(x)=x^{2}+2 x-8,$ being polynomial function, is continuous in $[-4,2]$ and is differentiable in $(-4,2).$

$f(-4)=(-4)^{2}+2 x(-4)-8=16-8-8=0$

$f(2)=(2)^{2}+2 \times 2-8=4+4-8=0$

$\therefore f(-4)=f(2)=0$

$\Rightarrow$ The value of $f(x)$ at $-4$ and $2$ coincides.

Rolle's Theorem states that there is a point $c \in(-4,2)$ such that $f^{\prime}(c)=0$

$f(x)=x^{2}+2 x-8$

$\Rightarrow f^{\prime}(x)=2 x+2$

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

$\Rightarrow 2 c+2=-1$

$\Rightarrow c=-1$

$c=-1 \in(-4,2)$

Hence, Rolle's Theorem is verified for the given function.