Answer
$c=-1$ or $c=1.5$
Work Step by Step
Here $g'(x)=3x^2 $ when $-2 \leq x \lt 0$; $g'(x)=0 $ when $x=0$ and $g'(x)=2x $ when $0 \lt x \leq 2$
The Mean value Theorem states that there is a point $c \in (-2,2)$
such that $f'(c)=\dfrac{g(2)-g(-2)}{2-(-2)}=\dfrac{4-(-8)}{4}=3$
This implies $3c^2=3$ when $x \lt 0 \implies c= \pm 1$
For $x \lt 0$, we have $c=-1$
and
For $x \gt 0$, we have $2c=3 \implies c=1.5$
Hence, $c=-1$ or $c=1.5$
