022 Exam 1 Sample A, Problem 4

Determine the intervals where the function  $h(x)=2x^{4}-x^{2}$ is increasing or decreasing.

Foundations:
When a first derivative is positive, the function is increasing (heading uphill). When the first derivative is negative, it is decreasing (heading downhill). When the first derivative is $0,$ it is not quite so clear. If $f'(z)=0$  at a point $z$, and the first derivative splits around it (either $f'(x)<0$  for $x<z$ and $f'(x)>0$  for $x>z$, or $f'(x)>0$  for $x<z$ and $f'(x)<0$  for $x>z$), then the point $(z,f(z))$ is a local maximum or minimum, respectively, and is neither increasing or decreasing at that point.
On the other hand, if the first derivative does not split around $z$, then it will be increasing or decreasing at that point based on the derivative of the adjacent intervals. For example, $g(x)=x^{3}$ has the derivative $g'(x)=3x^{2}$. Thus, $g'(0)=0$, but is strictly positive everywhere else. As a result, $g(x)=x^{3}$  is increasing on $(-\infty ,\infty )$.

 Solution:

Find the Roots of the First Derivative:
Note that
$h'(x)\,\,=\,\,8x^{3}-2x\,\,=\,\,2x\left(4x^{2}-1\right)\,\,=\,\,2x(2x+1)(2x-1),$
so the roots of $h'(x)$ are $0$  and $\pm 1/2$.

Make a Sign Chart and Evaluate:
We need to test convenient numbers on the intervals separated by the roots. Using the form
$h'(x)\,\,=\,\,2x(2x+1)(2x-1),$
we can test at convenient points to find
$f'(-10)=(-)(-)(-)=(-),\quad f'(-1/4)=(-)(+)(-)=(+),$
$f'(1/4)\,\,=(+)(+)(-)=(-),\quad f'(10)=(+)(+)(+)=(+).$
From this, we can build a sign chart:
$x:$ $x<-1/2$ $x=-1/2$ $-1/2<x<0$ $x=0$ $0<x<1/2$ $x=1/2$ $x>1/2$ $f'(x):$ $(-)$ $0$ $(+)$ $0$ $(-)$ $0$ $(+)$
Notice that at each of our roots, the derivative does split (changes sign as $x$ passes through each root of $h'(x)$), so the function is neither increasing or decreasing at each root. Thus, $h(x)$ is increasing on $(-1/2,0)\cup (1/2,\infty )$, and decreasing on $(-\infty ,-1/2)\cup (0,1/2)$.

Final Answer:
$h(x)$ is increasing on $(-1/2,0)\cup (1/2,\infty )$, and decreasing on $(-\infty ,-1/2)\cup (0,1/2)$.

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