009A Sample Final 1, Problem 2 - Revision history

Kayla Murray at 22:41, 2 December 2017

2017-12-02T22:41:02Z

Kayla Murray at 20:12, 18 March 2017

2017-03-18T20:12:34Z

Kayla Murray at 16:17, 27 February 2017

2017-02-27T16:17:32Z

Kayla Murray at 00:32, 26 February 2017

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Kayla Murray at 00:30, 26 February 2017

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Kayla Murray at 00:29, 26 February 2017

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Kayla Murray at 00:29, 26 February 2017

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Kayla Murray at 00:26, 26 February 2017

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Kayla Murray at 01:46, 19 February 2017

2017-02-19T01:46:32Z

Kayla Murray at 01:42, 19 February 2017

2017-02-19T01:42:51Z

@@ Line 12: / Line 12: @@
 <span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that &nbsp;<math style="vertical-align: -3px">f(x)</math>&nbsp; is differentiable at &nbsp;<math style="vertical-align: 0px">x=3</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
     \right.
 </math>
-<span class="exam">(a) Show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=3</math>.
+<span class="exam">(a) Show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 <span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that &nbsp;<math style="vertical-align: -3px">f(x)</math>&nbsp; is differentiable at &nbsp;<math style="vertical-align: 0px">x=3</math>.
@@ Line 71: / Line 71: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),</math>
 |-
-|<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous.
+|<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 |}
@@ Line 132: / Line 132: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous at <math style="vertical-align: 0px">x=3.</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>

@@ Line 8: / Line 8: @@
     \right.
 </math>
-<span class="exam">(a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3</math>.
+<span class="exam">(a) Show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=3</math>.
-<span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3</math>.
+<span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that &nbsp;<math style="vertical-align: -3px">f(x)</math>&nbsp; is differentiable at &nbsp;<math style="vertical-align: 0px">x=3</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|'''1.''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous at <math style="vertical-align: 0px">x=a</math>&thinsp; if <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
+|'''1.''' &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=a</math>&nbsp; if
 |-
-|'''2.''' The definition of derivative for <math style="vertical-align: -5px">f(x)</math>&thinsp; is &thinsp;<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
 |}
@@ Line 28: / Line 32: @@
 !Step 1: &nbsp;
 |-
-|We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math> We have
+|We first calculate &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math>&nbsp; We have
 |-
+|
@@ Line 43: / Line 47: @@
 !Step 2: &nbsp;
 |-
-|Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math> We have
+|Now, we calculate &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math>&nbsp; We have
 |-
+|
@@ Line 58: / Line 62: @@
 !Step 3: &nbsp;
 |-
-|Now, we calculate <math style="vertical-align: -5px">f(3).</math> We have
+|Now, we calculate &nbsp;<math style="vertical-align: -5px">f(3).</math>&nbsp; We have
 |-
+|
@@ Line 67: / Line 71: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),</math>
 |-
-|<math style="vertical-align: -5px">f(x)</math> is continuous.
+|<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous.
 |}
@@ Line 121: / Line 125: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
 |-
-|<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
+|<math style="vertical-align: -5px">f(x)</math>&nbsp; is differentiable at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 |}

@@ Line 67: / Line 67: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),</math>
 |-
-|<math>f(x)</math> is continuous.
+|<math style="vertical-align: -5px">f(x)</math> is continuous.
 |}

@@ Line 65: / Line 65: @@
 |Since
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),</math>
 |-
-|is continuous.
+|<math>f(x)</math> is continuous.
 |}

@@ Line 63: / Line 63: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>f(3)=4\sqrt{3+1}\,=\,8.</math>
 |-
-|Since <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
+|Since
 |}

@@ Line 113: / Line 113: @@
 !Step 3: &nbsp;
 |-
-|Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
+|Since
 |-
 |<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
@@ Line 127: / Line 129: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 14: / Line 14: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
 |'''1.''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous at <math style="vertical-align: 0px">x=a</math>&thinsp; if <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
@@ Line 33: / Line 31: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\
 &&\\
@@ Line 48: / Line 46: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\
 &&\\
@@ Line 63: / Line 61: @@
 |-
+|
-::<math>f(3)=4\sqrt{3+1}\,=\,8.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>f(3)=4\sqrt{3+1}\,=\,8.</math>
 |-
 |Since <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
@@ Line 76: / Line 74: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\
 &&\\
@@ Line 93: / Line 91: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4\sqrt{3+h+1}-8}{h}}\\
 &&\\
@@ Line 124: / Line 122: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
 |-
-|'''(b)''' Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
 |-
+|
-::<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the following piecewise defined function:
-::::::<math>f(x) = \left\{
+::<math>f(x) = \left\{
       \begin{array}{lr}
         x+5 &  \text{if }x < 3\\
@@ Line 8: / Line 8: @@
     \right.
 </math>
-<span class="exam">a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3</math>.
+<span class="exam">(a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3</math>.
-<span class="exam">b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3</math>.
+<span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"