Answer
a)
$f$ increasing on $(-1,0)\cup(1,\infty)$
$f$ decreasing on $(-\infty,-1)\cup(0,1)$
b)
$f(0) = 3$ is a local maximum value and
$f(-1) = 2, f(1) = 2$ are local minimum values.
c)
$f$ is concave upward on $\left(-\infty,-\frac{1}{\sqrt 3}\right)\cup\left(\frac{1}{\sqrt 3},\infty\right)$
$f$ is concave downward on $\left(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\right)$
There are inflection points at $\left(±\frac{1}{\sqrt 3},\frac{22}{9}\right)$
Work Step by Step
a)
$f(x) = x^{4}-2x^{2}+3$
$f'(x) = 4x^{3}-4x = 4x(x-1)(x+1)$
$f$ increasing on $(-1,0)\cup(1,\infty)$
$f$ decreasing on $(-\infty,-1\cup(0,1)$
b)
$f$ changes from increasing to decreasing at $x = 0$ and from decreasing to increasing at $x = -1, x = 1$. Thus,
$f(0) = 3$ is a local maximum value and
$f(-1) = 2, f(1) = 2$ are local minimum values.
c)
$f''(x) = 12x^{2}-4 = 12\left(x-\frac{1}{\sqrt 3}\right)\left(x+\frac{1}{\sqrt 3}\right)$
$f''(x) \gt 0$ for $x \lt -\frac{1}{\sqrt 3}$, $x \gt \frac{1}{\sqrt 3}$
$f''(x) \lt 0$ for $-\frac{1}{\sqrt 3} \lt x \lt \frac{1}{\sqrt 3}$
Thus
$f$ is concave upward on $\left(-\infty,-\frac{1}{\sqrt 3}\right)\cup\left(\frac{1}{\sqrt 3},\infty\right)$
$f$ is concave downward on $\left(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\right)$
There are inflection points at $\left(±\frac{1}{\sqrt 3},\frac{22}{9}\right)$
