## Calculus with Applications (10th Edition)

f(x) is decreasing on $(-\infty, -5) \cup (\frac{-1}{4},+\infty)$ and increasing on $(-5,\frac{-1}{4})$ On $(-\infty, -5)$, f(x) achieves a minimum value of $\frac{-377}{6}$ at x=−5 and on $(\frac{-1}{4},+\infty)$ a maximum value of $\frac{827}{96}$ at $x=\frac{-1}{4}$.
$f(x) =\frac{-4}{3}x^{3}-\frac{21}{2}x^{2}-5x+8$ $f'(x)=-4x^{2}-21x-5$ $f'(x)=0 \rightarrow -4x^{2}-21x-5=0 \rightarrow x=\frac{-1}{4}, x=-5$ Thus, f(x) is decreasing on $(-\infty, -5) \cup (\frac{-1}{4},+\infty)$ and increasing on $(-5,\frac{-1}{4})$ On $(-\infty, -5)$, f(x) achieves a minimum value of $\frac{-377}{6}$ at x=−5 and on $(\frac{-1}{4},+\infty)$ a maximum value of $\frac{827}{96}$ at $x=\frac{-1}{4}$.