Difference between revisions of "009A Sample Final 3, Problem 6"

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<span class="exam">(d) Sketch the shape of the graph of &nbsp;<math style="vertical-align: -4px">f.</math>
 
<span class="exam">(d) Sketch the shape of the graph of &nbsp;<math style="vertical-align: -4px">f.</math>
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[[009A Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Foundations: &nbsp;
 
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|-
 
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|'''1.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing when &nbsp;<math style="vertical-align: -5px">f'(x)>0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is decreasing when &nbsp;<math style="vertical-align: -5px">f'(x)<0.</math>
 
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|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
 
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|'''3.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up when &nbsp;<math style="vertical-align: -5px">f''(x)>0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave down when &nbsp;<math style="vertical-align: -5px">f''(x)<0.</math>
 
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[[009A Sample Final 3, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''Solution:'''
 
  
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'''(a)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 
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|We have &nbsp;<math style="vertical-align: -5px">f'(x)=24x^2-4x^3.</math>
 
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|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -6px">0=4x^2(6-x).</math>
 
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|Hence, we have &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=6.</math>
 
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|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,6),(6,\infty).</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 
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|For &nbsp;<math style="vertical-align: -5px">x=-1,~f'(x)=28>0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=1,~f'(x)=20>0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=7,~f'(x)=-196<0.</math>
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-\infty,6)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(6,\infty).</math>
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|The critical points of &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; occur at &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=6.</math>
 
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|Plugging these values into &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we get the critical points
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">(0,4)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(6,436).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|Using the first derivative test and the information from part (a),
 
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|&nbsp;<math style="vertical-align: -4px">(0,4)</math>&nbsp; is not a local minimum or local maximum and
 
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|&nbsp;<math style="vertical-align: -4px">(6,436)</math>&nbsp; is a local maximum.
 
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'''(c)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|To find the intervals when the function is concave up or concave down, we need to find &nbsp;<math style="vertical-align: -5px">f''(x).</math>
 
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|We have &nbsp;<math style="vertical-align: -5px">f''(x)=48x-12x^2.</math>
 
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|We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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|So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">0=12x(4-x).</math>&nbsp;
 
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|Hence, &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=4</math>.
 
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|This value breaks up the number line into three intervals:
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|Again, we use test points in these three intervals.
 
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|For &nbsp;<math style="vertical-align: -5px">x=-1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-60<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=48>0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=5,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-60<0.</math>
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on the interval &nbsp;<math style="vertical-align: -5px">(0,4)</math>&nbsp; and concave down on the interval &nbsp;<math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty).</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!(d): &nbsp;
 
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|Insert graph
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-\infty,6)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(6,\infty).</math>
 
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; The critical points are <math style="vertical-align: -4px">(0,4)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(6,436).</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">(0,4)</math>&nbsp; is not a local minimum or local maximum and &nbsp;<math style="vertical-align: -5px">(6,436)</math>&nbsp; is a local maximum.
 
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|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on the interval &nbsp;<math style="vertical-align: -5px">(0,4)</math>&nbsp; and concave down on the interval &nbsp;<math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty).</math>
 
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|&nbsp; &nbsp;'''(d)'''&nbsp; &nbsp; See above
 
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|}
 
 
[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 16:42, 2 December 2017

Let

(a) Over what  -intervals is    increasing/decreasing?

(b) Find all critical points of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}   and test each for local maximum and local minimum.

(c) Over what  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -intervals is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}   concave up/down?

(d) Sketch the shape of the graph of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f.}

Solution


Detailed Solution


Return to Sample Exam

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