Difference between revisions of "022 Exam 1 Sample A, Problem 4"

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|On the other hand, if the first derivative does not split around <math style="vertical-align: 0%">z</math>, then it will be increasing or decreasing at that point based on the derivative of the adjacent intervals.  For example, <math style="vertical-align: -25%">g(x)=x^3</math> has the derivative <math style="vertical-align: -20%">g'(x)=3x^2</math>.  Thus, <math style="vertical-align: -30%">g'(0)=0</math>, but is strictly positive every else.  As a result, <math style="vertical-align: -20%">g(x)=x^3</math> is increasing on <math style="vertical-align: -20%">(-\infty,\infty)</math>.
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|On the other hand, if the first derivative does not split around <math style="vertical-align: 0%">z</math>, then it will be increasing or decreasing at that point based on the derivative of the adjacent intervals.  For example, <math style="vertical-align: -25%">g(x)=x^3</math> has the derivative <math style="vertical-align: -20%">g'(x)=3x^2</math>.  Thus, <math style="vertical-align: -30%">g'(0)=0</math>, but is strictly positive every else.  As a result, <math style="vertical-align: -20%">g(x)=x^3</math>&thinsp; is increasing on <math style="vertical-align: -20%">(-\infty,\infty)</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Find the Derivatives and Their Roots: &nbsp;
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!Find the Roots of the First Derivative: &nbsp;
 
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|-
 
|Note that  
 
|Note that  
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::<math>h'(x)\,\,=\,\,8x^3-2x\,\,=\,\,2x\left(4x^2-1\right)\,\,=\,\,2x(2x+1)(2x-1),</math>
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|so the roots of <math style="vertical-align: -25%">h'(x)</math> are <math style="vertical-align: 0%">0</math>&thinsp; and <math style="vertical-align: -20%">\pm 1/2</math>.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Make a Sign Chart and Evaluate: &nbsp;
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|-
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|We need to test convenient numbers on the intervals separated by the roots.  Using the form
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|-
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|
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::<math>h'(x)\,\,=\,\,2x(2x+1)(2x-1),</math>
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|-
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|we can test at convenient points to find
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::<math>f'(-10)=(-)(-)(-)=(-),\quad f'(-1/4)=(-)(+)(-)=(+),</math>
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::<math>f'(1/4)\,\,=(+)(+)(-)=(-), \quad f'(10)=(+)(+)(+)=(+).</math>
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|From this, we can build a sign chart:<br>
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|-
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|<table border="1" cellspacing="0" cellpadding="6" align="center">
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  <tr>
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    <td align = "center"><math> x:</math></td>
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    <td align = "center"><math> x<-1/2 </math></td>
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    <td align = "center"><math> x=-1/2 </math></td>
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    <td align = "center"><math> -1/2<x<0 </math></td>
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    <td align = "center"><math> x=0</math></td>
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    <td align = "center"><math>0<x<1/2</math></td>
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    <td align = "center"><math> x=1/2</math></td>
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    <td align = "center"><math> x>1/2</math></td>
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  </tr>
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  <tr>
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    <td align = "center"><math> f'(x):</math></td>
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    <td align = "center"><math> (-) </math></td>
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    <td align = "center"><math> 0 </math></td>
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    <td align = "center"><math> (+) </math></td>
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    <td align = "center"><math> 0 </math></td>
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    <td align = "center"><math> (-)</math></td>
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    <td align = "center"><math> 0 </math></td>
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    <td align = "center"><math> (+) </math></td>
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  </tr>
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</table><br>
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|-
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|Notice that at each of our roots, the derivative does split (changes sign as <math style="vertical-align: 0%">x</math> passes through each root of <math style="vertical-align: -20%">h'(x)</math>), so the function is neither increasing or decreasing at each root.  Thus, <math style="vertical-align: -20%">h(x)</math> is increasing on <math style="vertical-align: -20%">(-1/2,0)\cup(1/2,\infty)</math>, and decreasing on <math style="vertical-align: -20%">(-\infty,-1/2)\cup(0,1/2)</math>.
 
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|}
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
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|-
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|<math style="vertical-align: -20%">h(x)</math> is increasing on <math style="vertical-align: -20%">(-1/2,0)\cup(1/2,\infty)</math>, and decreasing on <math style="vertical-align: -20%">(-\infty,-1/2)\cup(0,1/2)</math>.
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|}
 
[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 20:55, 12 April 2015

 Problem 4.  Determine the intervals where the function  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 , it is not quite so clear. If at a point , and the first derivative splits around it (either   for and   for , or   for and   for ), then the point 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 , then it will be increasing or decreasing at that point based on the derivative of the adjacent intervals. For example, has the derivative . Thus, , but is strictly positive every else. As a result,   is increasing on .

 Solution:

Find the Roots of the First Derivative:  
Note that
so the roots of are   and .
Make a Sign Chart and Evaluate:  
We need to test convenient numbers on the intervals separated by the roots. Using the form
we can test at convenient points to find
From this, we can build a sign chart:

Notice that at each of our roots, the derivative does split (changes sign as passes through each root of ), so the function is neither increasing or decreasing at each root. Thus, is increasing on , and decreasing on .
Final Answer:  
is increasing on , and decreasing on .

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