## Calculus with Applications (10th Edition)

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