Answer
a) Increasing on $(-2,0)\cup(2,\infty)$, decreasing on $(-\infty,-2)\cup(0,2)$
b) Local maximum value: $f(0)$ = $3$, local maximum values: $f(±2)$ = $-5$
c) Concave upward on $\left(-\infty,-\frac{2}{\sqrt 3}\right)\cup\left(\frac{2}{\sqrt 3},\infty\right)$, concave downward on $\left(-\frac{2}{\sqrt 3},\frac{2}{\sqrt 3}\right)$
Inflection points: $\left(±\frac{2}{\sqrt 3},-\frac{13}{9}\right)$
d) See graph
Work Step by Step
a)
$f(x)$ = $\frac{1}{2}x^{4}-4x^{2}+3$
$f'(x)$ = $2x^{3}-8x$ = $2x(x+2)(x-2)$
$f'(x)$ $\gt$ $0$ => $-2$ $\lt$ $x$ $\lt$ $0$ or $x$ $\gt$ $2$
$f'(x)$ $\lt$ $0\Rightarrow x$ $\lt$ $-2$ or $0$ $\lt$ $x$ $\lt$ $2$
so $f$ is increasing on $(-2,0)\cup(2,\infty)$ and
$f$ is decreasing on $(-\infty,-2)\cup(0,2)$
b)
$f$ changes from increasing to decreasing at $x$ = $0$ so $f(0)$ = $3$ is a local maximum value.
$f$ change from decreasing to increasing at $x$ = $±2$ so $f(±2)$ = $-5$ is a local minimum value
c)
$f''(x)$ = $6x^{2}-8$ = $6\left(x+\frac{2}{\sqrt 3}\right)\left(x-\frac{2}{\sqrt 3}\right)$
$f''(x)$ = $0$ => $x$ = $±\frac{2}{\sqrt 3}$
$f''(x)$ $\gt$ $0$ on $\left(-\infty,-\frac{2}{\sqrt 3}\right)\cup\left(\frac{2}{\sqrt 3},\infty\right)$
$f''(x)$ $\lt$ $0$ on $(-\frac{2}{\sqrt 3},\frac{2}{\sqrt 3})$
$f$ is concave upward on $\left(-\infty,-\frac{2}{\sqrt 3}\right)\cup\left(\frac{2}{\sqrt 3},\infty\right)$
$f$ is concave downward on $\left(-\frac{2}{\sqrt 3},\frac{2}{\sqrt 3}\right)$
There are inflection points at $\left(±\frac{2}{\sqrt 3},-\frac{13}{9}\right)$
d) See graph
