# 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 and $f'(x)>0$ for $x>z$ , or $f'(x)>0$ for $x 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 $x=0$ $0 $x=1/2$ $x>1/2$ $f'(x):$ $(-)$ $0$ $(+)$ $0$ $(-)$ $0$ $(+)$ 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 $(-\infty ,-1/2)\cup (0,1/2)$ .
$h(x)$ is increasing on $(-1/2,0)\cup (1/2,\infty )$ , and decreasing on $(-\infty ,-1/2)\cup (0,1/2)$ .