#### Answer

f is increasing in $(-\infty,-2) \cup (\frac{2}{3},+\infty)$ and decreasing in $(-2,\frac{2}{3})$

#### Work Step by Step

$f(x)=4x^{3}+8x^{2}-16x+11$
$f'(x)=12x^{2}+16x-16$
$f'(x)=0 \rightarrow 12x^{2}+16x-16=0 \rightarrow x=-2, x=\frac{2}{3}$
$x \lt -2 \rightarrow $ f is increasing
$-2\lt x \lt \frac{2}{3} \rightarrow $ f is decreasing
$x \gt \frac{2}{3} \rightarrow $f is increasing