# 022 Exam 1 Sample A, Problem 4

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Determine the intervals where the function  ${\displaystyle h(x)=2x^{4}-x^{2}}$ 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 ${\displaystyle 0,}$ it is not quite so clear. If ${\displaystyle f'(z)=0}$  at a point ${\displaystyle z}$, and the first derivative splits around it (either ${\displaystyle f'(x)<0}$  for ${\displaystyle x and ${\displaystyle f'(x)>0}$  for ${\displaystyle x>z}$, or ${\displaystyle f'(x)>0}$  for ${\displaystyle x and ${\displaystyle f'(x)<0}$  for $\displaystyle x> z$ ), then the point ${\displaystyle (z,f(z))}$ 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 ${\displaystyle z}$, then it will be increasing or decreasing at that point based on the derivative of the adjacent intervals. For example, ${\displaystyle g(x)=x^{3}}$ has the derivative ${\displaystyle g'(x)=3x^{2}}$. Thus, ${\displaystyle g'(0)=0}$, but is strictly positive everywhere else. As a result, ${\displaystyle g(x)=x^{3}}$  is increasing on ${\displaystyle (-\infty ,\infty )}$.

Solution:

Find the Roots of the First Derivative:
Note that
${\displaystyle h'(x)\,\,=\,\,8x^{3}-2x\,\,=\,\,2x\left(4x^{2}-1\right)\,\,=\,\,2x(2x+1)(2x-1),}$
so the roots of ${\displaystyle h'(x)}$ are ${\displaystyle 0}$  and ${\displaystyle \pm 1/2}$.
Make a Sign Chart and Evaluate:
We need to test convenient numbers on the intervals separated by the roots. Using the form
${\displaystyle h'(x)\,\,=\,\,2x(2x+1)(2x-1),}$
we can test at convenient points to find
${\displaystyle f'(-10)=(-)(-)(-)=(-),\quad f'(-1/4)=(-)(+)(-)=(+),}$
${\displaystyle f'(1/4)\,\,=(+)(+)(-)=(-),\quad f'(10)=(+)(+)(+)=(+).}$
From this, we can build a sign chart:
 ${\displaystyle x:}$ ${\displaystyle x<-1/2}$ ${\displaystyle x=-1/2}$ ${\displaystyle -1/2 ${\displaystyle x=0}$ ${\displaystyle 0 ${\displaystyle x=1/2}$ ${\displaystyle x>1/2}$ ${\displaystyle f'(x):}$ ${\displaystyle (-)}$ ${\displaystyle 0}$ ${\displaystyle (+)}$ ${\displaystyle 0}$ ${\displaystyle (-)}$ ${\displaystyle 0}$ ${\displaystyle (+)}$

Notice that at each of our roots, the derivative does split (changes sign as ${\displaystyle x}$ passes through each root of ${\displaystyle h'(x)}$), so the function is neither increasing or decreasing at each root. Thus, ${\displaystyle h(x)}$ is increasing on ${\displaystyle (-1/2,0)\cup (1/2,\infty )}$, and decreasing on ${\displaystyle (-\infty ,-1/2)\cup (0,1/2)}$.
${\displaystyle h(x)}$ is increasing on ${\displaystyle (-1/2,0)\cup (1/2,\infty )}$, and decreasing on ${\displaystyle (-\infty ,-1/2)\cup (0,1/2)}$.