009A Sample Final A, Problem 3 - Revision history

MathAdmin at 17:33, 12 April 2015

2015-04-12T17:33:11Z

MathAdmin at 17:13, 12 April 2015

2015-04-12T17:13:15Z

MathAdmin at 02:36, 12 April 2015

2015-04-12T02:36:53Z

MathAdmin at 16:15, 1 April 2015

2015-04-01T16:15:54Z

MathAdmin at 02:59, 30 March 2015

2015-03-30T02:59:09Z

MathAdmin at 02:48, 30 March 2015

2015-03-30T02:48:31Z

MathAdmin at 18:53, 28 March 2015

2015-03-28T18:53:03Z

MathAdmin at 18:52, 28 March 2015

2015-03-28T18:52:44Z

MathAdmin at 18:51, 28 March 2015

2015-03-28T18:51:53Z

MathAdmin at 13:48, 28 March 2015

2015-03-28T13:48:39Z

@@ Line 35: / Line 35: @@
 |A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14%">x_0 </math> if
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow x_0} f(x) = f\left(x_0\right).</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow x_{_0}} f(x) = f\left(x_0\right).</math>
 |-
 |This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at <math style="vertical-align: -14%">x_0</math>. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.

@@ Line 37: / Line 37: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow x_0} f(x) = f\left(x_0\right).</math>
 |-
-|This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at<math style="vertical-align: -12%">x_0</math>. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.
+|This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at <math style="vertical-align: -14%">x_0</math>. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.
 |-
 |When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right.  This means we must show the derivatives agree for both the descriptions of <math style="vertical-align: -20%;">f</math> at the transition point.<br>

@@ Line 1: / Line 1: @@
 [[File:009A_SF_A_3.png|right|230px|frame|Both functions with constants chosen to provide continuity.]]
-<span style="font-size:135%"><font face=Times Roman>3. (Version I) Consider the following function:
+<span class="exam">3. (Version I) Consider the following function:
 &nbsp;<math style="vertical-align: -80%;">f(x) = \begin{cases} \sqrt{x}, & \mbox{if }x\geq 1, \\ 4x^{2}+C, & \mbox{if }x<1. \end{cases}</math>
 <br>
-<table border="0" style="font-size:135%">
+<table border="0" class="exam">
 <tr style="vertical-align:top">
   <td>&nbsp;(a) </td>
@@ Line 13: / Line 13: @@
 </table>
-<span style="font-size:135%">3. (Version II) Consider the following function:
+<span class="exam">3. (Version II) Consider the following function:
 &nbsp;<math style="vertical-align: -80%;">g(x)=\begin{cases}
 \sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\
 \frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1.
 \end{cases}</math>
-<table border="0" style="font-size:135%">
+<table border="0" class="exam">
 <tr style="vertical-align:top">
   <td>&nbsp;(a) </td>

@@ Line 33: / Line 33: @@
 ! Foundations: &nbsp;
 |-
-|A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14%;">x_0 </math> if
+|A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14%">x_0 </math> if
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow x_0} f(x) = f\left(x_0\right).</math>
 |-
-|This can be viewed as saying the left and right hand limits exist, and are equal. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.
+|This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at<math style="vertical-align: -12%">x_0</math>. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.
 |-
 |When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right.  This means we must show the derivatives agree for both the descriptions of <math style="vertical-align: -20%;">f</math> at the transition point.<br>

@@ Line 6: / Line 6: @@
 <table border="0" style="font-size:135%">
 <tr style="vertical-align:top">
-  <td>(a) </td>
+  <td>&nbsp;(a) </td>
   <td>Find a value of &nbsp;<math style="vertical-align: -0.1%;">C</math> which makes <math>f</math> continuous at <math style="vertical-align: -3%;">x=1.</math></td>
 <tr style="vertical-align:top">
-  <td>(b)</td>
+  <td>&nbsp;(b)</td>
 <td>With your choice of &nbsp;<math style="vertical-align: -0.1%;">C</math>, is <math>f</math> differentiable at <math style="vertical-align: -3%;">x=1</math>? &nbsp;Use the definition of the derivative to motivate your answer.</td>
 </table>
@@ Line 20: / Line 20: @@
 <table border="0" style="font-size:135%">
 <tr style="vertical-align:top">
-  <td>(a) </td>
+  <td>&nbsp;(a) </td>
   <td>Find a value of &nbsp;<math style="vertical-align: -0.1%;">C</math> which makes <math>f</math> continuous at <math style="vertical-align: -2.95%;">x=1.</math>
 </td>
 <tr style="vertical-align:top">
-  <td>(b)</td>
+  <td>&nbsp;(b)</td>
   <td>With your choice of &nbsp;<math style="vertical-align: -0.1%;">C</math>, is <math>f</math> differentiable at <math style="vertical-align: -3%;">x=1</math>? &nbsp;Use the definition of the derivative to motivate your answer.</td>
 </table>

@@ Line 85: / Line 85: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{2}.</math>
 |-
-|Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>, and <math style="vertical-align: -14%;">g(x)</math><br> is differentiable at the required point.
+|Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>, and <math style="vertical-align: -15%;">g(x)</math><br> is differentiable at the required point.
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 67: / Line 67: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,</math>
 |-
-|so <math style="vertical-align: -24%;">C=-7/4</math> makes the function continuous.
+|so <math style="vertical-align: -24%;">C=7/4</math> makes the function continuous.
 |-
 |(b) We again consider the derivative from each side of 1. For <math style="vertical-align: -2%;">x>1</math>,
@@ Line 85: / Line 85: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{2}.</math>
 |-
-|Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>.<br>
+|Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>, and <math style="vertical-align: -13%;">g(x)</math><br> is differentiable at the required point.
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]