Answer
Local minimum:$\quad (-2,0)$ and $(-2,0)$
Local maximum:$\quad$ $(0,4)$
Inflection points:$\quad (\displaystyle \pm\frac{2}{\sqrt{3}},\frac{16}{9})$
Intervals where f is concave down:$\quad (-\displaystyle \frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}})$
Intervals where f is concave up:$\quad (-\displaystyle \infty,-\frac{2}{\sqrt{3}})\cup(\frac{2}{\sqrt{3}},\infty)$
Work Step by Step
$y=f(x)=\displaystyle \frac{x^{4}}{4}-2x^{2}+4$
$f'(x)=x^{3}-4x=x(x^{2}-4)=x(x-2)(x+2)$,
which is defined everywhere.
Critical points: zeros of $f'$...$\quad x=-2,0,2$
$f''(x)=3x^{2}-4$
$f''(x)=0$ for $x=\displaystyle \pm\frac{2}{\sqrt{3}}\Rightarrow(\pm\frac{2}{\sqrt{3}}, f(\displaystyle \pm\frac{2}{\sqrt{3}}))$ are points of inflection
$f(\displaystyle \pm\frac{2}{\sqrt{3}})=\frac{16}{9}$
Using $\cap$ for "concave down" and $\cup$ for "concave up"
$\left[\begin{array}{llll}
interval & (-\infty,-\frac{2}{\sqrt{3}}) & (-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}) & (\frac{2}{\sqrt{3}},\infty)\\
t & -2 & 0 & 2\\
f''(t) & 8\gt 0 & -4\lt 0 & 8\gt 0\\
concavity & \cup & \cap & \cup
\end{array}\right]$
Apply the second derivative test (Th. 5)$\quad c=-2,0,2$
$\left\{\begin{array}{lll}
-2\in(-\infty,-\frac{2}{\sqrt{3}}) & \Rightarrow f''(-2)\gt 0\Rightarrow\min,, & f(-2)=0\\
0\in(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}) & \Rightarrow f''(0)\lt 0\Rightarrow\max, & f(0)=4\\
2\in(\frac{2}{\sqrt{3}},\infty) & f''(2)\gt 0\Rightarrow\min, & f(2)=0
\end{array}\right.$
