Step 1:

Now $f(x)=x^2+2x-8$ is a polynomial.

$\Rightarrow$ It is continuous and derivable in its domain $x\in R$.Hence it is continuous in the interval $[4,2]$ and derivable in the interval $[-4,2]$

$f(-4)=(-4)^2+2(-4)-8$

$\qquad\;\;=16-8-8$

$\qquad\;\;=0$

$f(2)=2^2+2(2)-8$

$\qquad\;\;=4+4-8$

$\qquad\;\;=0$

Step 2:

$f(x)=2x+2$

$\quad\;\;=2c+2=0$

$2c=-2$

$c=-1$

$f'(c)=0$ at $c=-1$

Hence the condition of Rolle's theorem is satisfied