Answer
f(x) is increasing on $(-\infty, \frac{1}{3})$ and decreasing on $(\frac{1}{3},+\infty)$.
On $(-\infty,+\infty)$ f(x) achieves a maximum value of -14/3 at x=1/3
Work Step by Step
$f(x) =-3x^{2}+2x-5$
$f'(x)=-6x+2$
$f'(x)=0 \rightarrow -6x+2=0 \rightarrow x=\frac{1}{3}$
Thus, f(x) is increasing on $(-\infty, \frac{1}{3})$ and decreasing on $(\frac{1}{3},+\infty)$
On $(-\infty,+\infty)$ f(x) achieves a maximum value of -14/3 at x=1/3
