009A Sample Final 2, Problem 10 - Revision history

Kayla Murray at 17:29, 1 December 2017

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Kayla Murray at 00:25, 21 May 2017

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Kayla Murray at 20:35, 18 March 2017

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Kayla Murray at 03:13, 8 March 2017

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Kayla Murray at 02:47, 8 March 2017

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Kayla Murray at 02:36, 8 March 2017

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Kayla Murray at 02:11, 8 March 2017

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Kayla Murray at 21:55, 6 March 2017

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Kayla Murray at 01:57, 19 February 2017

2017-02-19T01:57:20Z

@@ Line 152: / Line 152: @@
 !Step 1: &nbsp;
 |-
-|First, we note that the degree of the numerator is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and
+|By L'Hopital's Rule, we have
 |-
-|the degree of the denominator is &nbsp;<math style="vertical-align: 0px">2.</math>&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 160: / Line 176: @@
 !Step 2: &nbsp;
 |-
-|Since the degree of the denominator is greater than the degree of the numerator,
+|Since
 |-
 |<math style="vertical-align: -5px">f(x)</math>&nbsp; has a horizontal asymptote

@@ Line 44: / Line 44: @@
 |Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp;
 |-
-|So, we have &nbsp;<math style="vertical-align: -6px">-4(x^2-1)=0.</math>
+|So, we have
 |-
 |Hence, we have &nbsp;<math style="vertical-align: 0px">x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=-1.</math>
@@ Line 104: / Line 106: @@
 |We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |-
-|So, we have &nbsp;<math style="vertical-align: -1px">8x(x^2-3)=0.</math>&nbsp;
+|So, we have
 |-
 |Hence,
@@ Line 112: / Line 116: @@
 |This value breaks up the number line into four intervals:
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3}),(-\sqrt{3},0)(0,\sqrt{3}),(\sqrt{3},\infty).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3}),(-\sqrt{3},0),(0,\sqrt{3}),(\sqrt{3},\infty).</math>
 |}

@@ Line 86: / Line 86: @@
 |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)=6x-12.</math>
+|Using the Quotient Rule and Chain Rule, we have
 |-
 |We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |-
-|So, we have &nbsp;<math style="vertical-align: -1px">0=6x-12.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: 0px">x=2.</math>
+|So, we have &nbsp;<math style="vertical-align: -1px">8x(x^2-3)=0.</math>&nbsp;
 |-
-|This value breaks up the number line into two intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
+|Hence,
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
 |}
@@ Line 100: / Line 118: @@
 !Step 2: &nbsp;
 |-
+|Again, we use test points in these four intervals.
 |}
@@ Line 143: / Line 183: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;The local maximum value of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">2</math>&nbsp; and the local minimum value of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">-2</math>&nbsp;
 |-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on &nbsp;<math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>&nbsp; and concave down on &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
 |-
 |&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>y=0</math>

@@ Line 84: / Line 84: @@
 !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>
 |-
+|

@@ Line 30: / Line 30: @@
 !Step 1: &nbsp;
 |-
+|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 |-
+|Hence, we have &nbsp;<math style="vertical-align: 0px">x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=-1.</math>
 |-
+|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-1),(-1,1),(1,\infty).</math>
 |}
@@ Line 42: / Line 56: @@
 !Step 2: &nbsp;
 |-
+|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 |-
+|
@@ Line 99: / Line 131: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-1,1)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;
 |-
 |&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>y=0</math>

@@ Line 13: / Line 13: @@
 !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>
 |-
+|'''4.''' Inflection points occur when &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |}
@@ Line 94: / Line 94: @@
 !(d): &nbsp;
 |-
+|Insert sketch
 |-
+|

@@ Line 2: / Line 2: @@
 ::<math>f(x)=\frac{4x}{x^2+1}</math>
-<span class="exam">(a) Find all local maximum and local minimum values of <math>f,</math> find all intervals where <math>f</math> is increasing and all intervals where <math>f</math> is decreasing.
+<span class="exam">(a) Find all local maximum and local minimum values of &nbsp;<math style="vertical-align: -4px">f,</math>&nbsp; find all intervals where &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is increasing and all intervals where &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is decreasing.
-<span class="exam">(b) Find all inflection points of the function <math>f,</math> find all intervals where the function <math>f</math> is concave upward and all intervals where <math>f</math> is concave downward.
+<span class="exam">(b) Find all inflection points of the function &nbsp;<math style="vertical-align: -4px">f,</math>&nbsp; find all intervals where the function &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is concave upward and all intervals where &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is concave downward.
-<span class="exam">(c) Find all horizontal asymptotes of the graph <math>y=f(x).</math>
+<span class="exam">(c) Find all horizontal asymptotes of the graph &nbsp;<math style="vertical-align: -5px">y=f(x).</math>
-<span class="exam">(d) Sketch the graph of <math>y=f(x).</math>
+<span class="exam">(d) Sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x).</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 81: / Line 81: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
@@ Line 100: / Line 112: @@
 |-
 |'''(c)'''
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]