Verify Rolle’s Theorem for the function
Question:

Verify Rolle’s Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$

Solution:

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

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

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

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

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

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

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

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

$\Rightarrow 2 c+2=0$

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

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