Answer
Local minimum: $(2,-3)$
Local maximum: $(-1,1.5)$
Inflection points: $(0.5,\ -0.75)$
Intervals where f is concave down: $(-\infty,0.5)$
Intervals where f is concave up: $(0.5,\infty)$
Work Step by Step
$y=f(x)=\displaystyle \frac{x^{3}}{3}-\frac{x^{2}}{2}-2x+\frac{1}{3}$
$f'(x)=x^{2}-x-2=(x-2)(x+1)$,
which is defined everywhere,
Critical points: zeros of $f'$...$\quad x=-1,2$
$f''(x)=0$ for $x=\displaystyle \frac{1}{2}\Rightarrow(\frac{1}{2}, f(\displaystyle \frac{1}{2}))$ is the point of inflection, (concavity changes)
$f(0.5)=-0.75$
Using $\cap$ for "concave down" and $\cup$ for "concave up"
$\left[\begin{array}{lll}
interval & (-\infty,0.5) & (0.5,\infty)\\
t & 0 & 1\\
f''(t) & -1\lt 0 & 1\gt 0\\
concavity & \cap & \cup
\end{array}\right]$
Apply the second derivative test (Th. 5)
$\left\{\begin{array}{lll}
-1\in(-\infty,0.5) & \Rightarrow f''(-1)\lt 0\Rightarrow\max, & f(-1)=1.5\\
& & \\
2\in(0.5,\infty) & f''(2)\gt 0\Rightarrow\min, & f(2)=-3
\end{array}\right.$
