Answer
(a) $f$ is decreasing on the intervals $(-\infty,-2)\cup(0,2)$
$f$ is increasing on the intervals $(-2,0)\cup (2,\infty)$
(b) The local maximum is $f(0) = 3$
The local minima are $f(-2) = f(2) = -5$
(c) The graph is concave down on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$
The graph is concave up on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3}, \infty)$
The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{13}{9})$ and $(\frac{2\sqrt{3}}{3}, -\frac{13}{9})$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = \frac{1}{2}x^4-4x^2+3$
We can find the points where $f'(x) = 0$:
$f'(x) = 2x^3-8x = 0$
$2x(x^2-4) = 0$
$2x(x-2)(x+2) = 0$
$x = -2,0,2$
When $x \lt -2~~$ or $0 \lt x \lt 2~~$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(-\infty,-2)\cup(0,2)$
When $-2 \lt x \lt 0$ or $x \gt 2$ then $f'(x) \gt 0$
$f$ is increasing on the intervals $(-2,0)\cup (2,\infty)$
(b) $f(-2) = \frac{1}{2}(-2)^4-4(-2)^2+3 = -5$
$f(0) = \frac{1}{2}(0)^4-4(0)^2+3 = 3$
$f(2) = \frac{1}{2}(2)^4-4(2)^2+3 = -5$
The local maximum is $f(0) = 3$
The local minima are $f(-2) = f(2) = -5$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 6x^2-8= 0$
$3x^2-4 = 0$
$x^2 = \frac{4}{3}$
$x = \pm \sqrt{\frac{4}{3}}$
$x = -\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3}, \infty)$
$f(-\frac{2\sqrt{3}}{3}) = \frac{1}{2}(-\frac{2\sqrt{3}}{3})^4-4(-\frac{2\sqrt{3}}{3})^2+3 = -\frac{13}{9}$
$f(\frac{2\sqrt{3}}{3}) = \frac{1}{2}(\frac{2\sqrt{3}}{3})^4-4(\frac{2\sqrt{3}}{3})^2+3 = -\frac{13}{9}$
The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{13}{9})$ and $(\frac{2\sqrt{3}}{3}, -\frac{13}{9})$
(d) We can see a sketch of the graph below.