009A Sample Final 1, Problem 10 - Revision history

Kayla Murray at 23:17, 2 December 2017

2017-12-02T23:17:04Z

Kayla Murray at 20:26, 18 March 2017

2017-03-18T20:26:03Z

Kayla Murray at 19:18, 7 March 2017

2017-03-07T19:18:44Z

Kayla Murray at 16:41, 27 February 2017

2017-02-27T16:41:51Z

Kayla Murray at 01:08, 26 February 2017

2017-02-26T01:08:04Z

Kayla Murray at 01:49, 19 February 2017

2017-02-19T01:49:58Z

Kayla Murray at 01:45, 19 February 2017

2017-02-19T01:45:25Z

Grad at 19:24, 3 March 2016

2016-03-03T19:24:16Z

Grad: /* Temp2 */

2016-03-03T19:23:54Z

Temp2

Grad: /* Temp2 */

2016-03-03T19:23:09Z

Temp2

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the following continuous function:
+<span class="exam">If a resistor of &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; ohms is connected across a battery of &nbsp;<math style="vertical-align: 0px">E</math>&nbsp; volts with internal resistance &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; ohms, then the power (in watts) in the external resistor is
-<span class="exam">defined on the closed, bounded interval &nbsp;<math style="vertical-align: -5px">[-8,8]</math>.
+::<math>P=\frac{E^2R}{(R+r)^2}.</math>
-<span class="exam">(a) Find all the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>.
+<span class="exam">If &nbsp;<math style="vertical-align: 0px">E</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; are fixed but &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; varies, what is the maximum value of the power?
-<span class="exam">(b) Determine the absolute maximum and absolute minimum values for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on the interval &nbsp;<math style="vertical-align: -5px">[-8,8]</math>.
+<hr>
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 58: / Line 58: @@
 |Solving, we get &nbsp;<math style="vertical-align: -1px">x=2.</math>
 |-
-|Thus, the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; are &nbsp;<math style="vertical-align: -5px">(0,0)</math> and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
+|Thus, the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; are &nbsp;<math style="vertical-align: -5px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
 |}

@@ Line 85: / Line 85: @@
 |&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: -4px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
 |-
-|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; The absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; The absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">32</math>&nbsp; and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 ::<math>f(x)=x^{1/3}(x-8)</math>
-<span class="exam">defined on the closed, bounded interval <math style="vertical-align: -5px">[-8,8]</math>.
+<span class="exam">defined on the closed, bounded interval &nbsp;<math style="vertical-align: -5px">[-8,8]</math>.
-<span class="exam">(a) Find all the critical points for <math style="vertical-align: -5px">f(x)</math>.
+<span class="exam">(a) Find all the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>.
-<span class="exam">(b) Determine the absolute maximum and absolute minimum values for <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[-8,8]</math>.
+<span class="exam">(b) Determine the absolute maximum and absolute minimum values for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on the interval &nbsp;<math style="vertical-align: -5px">[-8,8]</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|'''1.''' To find the critical points for <math style="vertical-align: -5px">f(x),</math> we set <math style="vertical-align: -5px">f'(x)=0</math> and solve for <math style="vertical-align: -1px">x.</math>
+|'''1.''' To find the critical points for &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we set &nbsp;<math style="vertical-align: -5px">f'(x)=0</math>&nbsp; and solve for &nbsp;<math style="vertical-align: -1px">x.</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
+&nbsp; &nbsp; &nbsp; &nbsp; Also, we include the values of &nbsp;<math style="vertical-align: -1px">x</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f'(x)</math>&nbsp; is undefined.
 |-
-|'''2.''' To find the absolute maximum and minimum of <math style="vertical-align: -5px">f(x)</math> on an interval <math>[a,b],</math>
+|'''2.''' To find the absolute maximum and minimum of &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on an interval &nbsp;<math>[a,b],</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
+&nbsp; &nbsp; &nbsp; &nbsp; we need to compare the &nbsp;<math style="vertical-align: -5px">y</math>&nbsp; values of our critical points with &nbsp;<math style="vertical-align: -5px">f(a)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(b).</math>
 |}
@@ Line 30: / Line 30: @@
 !Step 1: &nbsp;
 |-
-|To find the critical points, first we need to find <math style="vertical-align: -5px">f'(x).</math>
+|To find the critical points, first we need to find &nbsp;<math style="vertical-align: -5px">f'(x).</math>
 |-
 |Using the Product Rule, we have
@@ Line 45: / Line 45: @@
 !Step 2: &nbsp;
 |-
-|Notice <math style="vertical-align: -5px">f'(x)</math> is undefined when <math style="vertical-align: -1px">x=0.</math>
+|Notice &nbsp;<math style="vertical-align: -5px">f'(x)</math>&nbsp; is undefined when &nbsp;<math style="vertical-align: -1px">x=0.</math>
 |-
-|Now, we need to set <math style="vertical-align: -5px">f'(x)=0.</math>
+|Now, we need to set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>
 |-
 |So, we get
@@ Line 54: / Line 54: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
 |-
-|We cross multiply to get <math style="vertical-align: 1px">-3x=x-8.</math>
+|We cross multiply to get &nbsp;<math style="vertical-align: 1px">-3x=x-8.</math>
 |-
-|Solving, we get <math style="vertical-align: -1px">x=2.</math>
+|Solving, we get &nbsp;<math style="vertical-align: -1px">x=2.</math>
 |-
-|Thus, the critical points for <math style="vertical-align: -5px">f(x)</math> are <math style="vertical-align: -5px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
+|Thus, the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; are &nbsp;<math style="vertical-align: -5px">(0,0)</math> and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
 |}
@@ Line 66: / Line 66: @@
 !Step 1: &nbsp;
 |-
-|We need to compare the values of <math style="vertical-align: -5px">f(x)</math>&thinsp; at the critical points and at the endpoints of the interval.
+|We need to compare the values of &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the critical points and at the endpoints of the interval.
 |-
-|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math>&thinsp; and <math style="vertical-align: -5px">f(8)=0.</math>
+|Using the equation given, we have &nbsp;<math style="vertical-align: -5px">f(-8)=32</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(8)=0.</math>
 |}
@@ Line 74: / Line 74: @@
 !Step 2: &nbsp;
 |-
-|Comparing the values in Step 1 with the critical points in '''(a)''', the absolute maximum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -1px">32</math>
+|Comparing the values in Step 1 with the critical points in (a), the absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">32</math>
 |-
-|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
@@ Line 83: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: -4px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
 |-
-|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; The absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
 |'''1.''' To find the critical points for <math style="vertical-align: -5px">f(x),</math> we set <math style="vertical-align: -5px">f'(x)=0</math> and solve for <math style="vertical-align: -1px">x.</math>
 |-
+|
-::Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
+&nbsp; &nbsp; &nbsp; &nbsp; Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
 |-
 |'''2.''' To find the absolute maximum and minimum of <math style="vertical-align: -5px">f(x)</math> on an interval <math>[a,b],</math>
 |-
+|
-::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
+&nbsp; &nbsp; &nbsp; &nbsp; we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
 |}
@@ Line 37: / Line 35: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
 &&\\
@@ Line 54: / Line 52: @@
 |-
+|
-::<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
 |-
 |We cross multiply to get <math style="vertical-align: 1px">-3x=x-8.</math>
@@ Line 85: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''&thinsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
 |-
-|'''(b)'''&thinsp; The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 '''Solution:'''
-== Temp1 ==
 '''(a)'''
@@ Line 62: / Line 62: @@
 |}
 '''(b)'''
@@ Line 81: / Line 80: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;

@@ Line 68: / Line 68: @@
 !Step 1: &nbsp;
 |-
-|We need to compare the values of <math style="vertical-align: -5px">f(x)</math> at the critical points and at the endpoints of the interval.
+|We need to compare the values of <math style="vertical-align: -5px">f(x)</math>&thinsp; at the critical points and at the endpoints of the interval.
 |-
-|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math> and <math style="vertical-align: -5px">f(8)=0.</math>
+|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math>&thinsp; and <math style="vertical-align: -5px">f(8)=0.</math>
 |}
@@ Line 78: / Line 78: @@
 |Comparing the values in Step 1 with the critical points in '''(a)''', the absolute maximum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -1px">32</math>
 |-
-|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
 == Temp3 ==
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"