Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 197: 6

$c=-1$ or $c=1.5$

Since, the function $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 $c$ and $c \in (-2,2)$ such that $f'(c)=\dfrac{g(2)-g(-2)}{[2-(-2)]}=\dfrac{4-(-8)}{4}=3$ So, $3c^2=3$ when $x \lt 0 $ and $c= \pm 1$ Now, for $x \lt 0$, we have $c=-1$; & $x \gt 0$, we have $2c=3$ or, $ c=1.5$ Thus, $c=-1$ or $c=1.5$