009A Sample Final 3, Problem 6 - Revision history

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← Older revision		Revision as of 23:42, 2 December 2017
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	<span class="exam">(d) Sketch the shape of the graph of  <math style="vertical-align: -4px">f.</math>		<span class="exam">(d) Sketch the shape of the graph of  <math style="vertical-align: -4px">f.</math>
		+	<hr>
		+	[[009A Sample Final 3, Problem 6 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	\|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when  <math style="vertical-align: -5px">f'(x)>0</math>  and  <math style="vertical-align: -5px">f(x)</math>  is decreasing when  <math style="vertical-align: -5px">f'(x)<0.</math>
−	\|-
−	~~\|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.~~
−	\|-
−	\|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when  <math style="vertical-align: -5px">f''(x)>0</math>  and  <math style="vertical-align: -5px">f(x)</math>  is concave down when  <math style="vertical-align: -5px">f''(x)<0.</math>
−	\|}

		+	[[009A Sample Final 3, Problem 6 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

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

@@ Line 29: / Line 29: @@
 !Step 1: &nbsp;
 |-
-|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp; We have &nbsp;<math style="vertical-align: -5px">f'(x)=24x^2-4x^3.</math>
+|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 |-
-|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have &nbsp;<math style="vertical-align: -6px">0=4x^2(6-x).</math>
+|We have &nbsp;<math style="vertical-align: -5px">f'(x)=24x^2-4x^3.</math>
 |-
 |Hence, we have &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=6.</math>
 |-
-|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,6),(6,\infty).</math>
+|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 |}
@@ Line 87: / Line 93: @@
 |We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |-
-|So, we have &nbsp;<math style="vertical-align: -1px">0=12x(4-x).</math>&nbsp; Hence, &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=4</math>.
+|So, we have
 |-
-|This value breaks up the number line into three intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
 |}

@@ Line 81: / Line 81: @@
 !Step 1: &nbsp;
 |-
+|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>
 |-
+|We have &nbsp;<math style="vertical-align: -5px">f''(x)=48x-12x^2.</math>
 |-
+|We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |-
+|So, we have &nbsp;<math style="vertical-align: -1px">0=12x(4-x).</math>&nbsp; Hence, &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=4</math>.
 |-
+|This value breaks up the number line into three intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
 |}
@@ Line 95: / Line 95: @@
 !Step 2: &nbsp;
 |-
+|Again, we use test points in these three intervals.
 |-
+|For &nbsp;<math style="vertical-align: -5px">x=-1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-60<0.</math>
 |-
+|For &nbsp;<math style="vertical-align: -5px">x=1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=48>0.</math>
 |-
+|For &nbsp;<math style="vertical-align: -5px">x=5,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-60<0.</math>
 |}
@@ Line 120: / Line 122: @@
 |&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.
 |-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;
+|&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>
 |-
 |&nbsp; &nbsp;'''(d)'''&nbsp; &nbsp; See above
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 59: / Line 59: @@
 !Step 1: &nbsp;
 |-
+|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>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">(0,4)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(6,436).</math>
 |}
@@ Line 67: / Line 69: @@
 !Step 2: &nbsp;
 |-
+|Using the first derivative test and the information from part (a),
 |}
@@ Line 110: / Line 116: @@
 |&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>
 |-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;
+|&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>
 |-
 |&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;

@@ Line 41: / Line 41: @@
 !Step 2: &nbsp;
 |-
+|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 |-
+|
@@ Line 95: / Line 103: @@
 |Insert graph
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;
+|&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>
 |-
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;

@@ Line 14: / Line 14: @@
 !Foundations: &nbsp;
 |-
+|'''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>
 |-
+|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
 |-
+|'''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>
 |}

@@ Line 92: / Line 92: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 101: / Line 114: @@
 |-
 |'''(c)'''
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::<math>f(x)=4+8x^3-x^4</math>
-<span class="exam">(a) Over what <math>x</math>-intervals is <math>f</math> increasing/decreasing?
+<span class="exam">(a) Over what &nbsp;<math style="vertical-align: 0px">x</math>-intervals is &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; increasing/decreasing?
-<span class="exam">(b) Find all critical points of <math>f</math> and test each for local maximum and local minimum.
+<span class="exam">(b) Find all critical points of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; and test each for local maximum and local minimum.
-<span class="exam">(c) Over what <math>x</math>-intervals is <math>f</math> concave up/down?
+<span class="exam">(c) Over what &nbsp;<math style="vertical-align: 0px">x</math>-intervals is &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; concave up/down?
-<span class="exam">(d) Sketch the shape of the graph of <math>f.</math>
+<span class="exam">(d) Sketch the shape of the graph of &nbsp;<math style="vertical-align: -4px">f.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"