022 Exam 1 Sample A, Problem 1 - Revision history

MathAdmin at 05:34, 12 April 2015

2015-04-12T05:34:19Z

MathAdmin at 19:43, 2 April 2015

2015-04-02T19:43:36Z

MathAdmin at 18:36, 2 April 2015

2015-04-02T18:36:13Z

MathAdmin at 16:39, 2 April 2015

2015-04-02T16:39:51Z

MathAdmin at 05:20, 2 April 2015

2015-04-02T05:20:23Z

MathAdmin at 05:19, 2 April 2015

2015-04-02T05:19:19Z

MathAdmin: Created page with "1. Use the definition of derivative to find the derivative of $f(x)=\sqrt{x-5}$ .
2015-04-02T05:17:20Z

Created page with "<span style="font-size:135%"><font face=Times Roman>1. Use the definition of derivative to find the derivative of <math style="vertical-align: -15%">f(x)=\sqrt{x-5}</math>.</f..."

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<span style="font-size:135%"><font face=Times Roman>1. Use the definition of derivative to find the derivative
of <math style="vertical-align: -15%">f(x)=\sqrt{x-5}</math>.</font face=Times Roman>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall that the derivative is actually defined through the limit
|-
|     <math>f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}.</math>
|-
|The goal in solving this is to plug the values <math style="vertical-align: -10%">x+h</math> and <math style="vertical-align: 0%">x</math> into the appropriate <math style="vertical-align: -20%">f</math> in the numerator, and then find a way to cancel the <math style="vertical-align: 0%">h</math> in the denominator. Unlike simplifying a rational expression containing radicals, here it's appropriate to have a radical in the denominator.
|-
|Again, the goal is to cancel the <math style="vertical-align: 0%">h</math>.
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Following the hints above, we initially have
|-
|     <math>f'(x)\,=\,\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\,=\,\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We can't plug in <math style="vertical-align: 0%">h=0</math>, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:
<table>
<tr>
<td>     <math>\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}\cdot\frac{\sqrt{x+h-5}+\sqrt{x-5}}{\sqrt{x+h-5}+\sqrt{x-5}}</math></td>
<td> <math>=\lim_{h\rightarrow0}\frac{\left(\sqrt{x+h-5}\right)^{2}-\left(\sqrt{x-5}\right)^{2}}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
</tr>
<tr>
<td></td>
<td> <math>=\lim_{h\rightarrow0}\frac{x+h-5-\left(x-5\right)}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math> </td>
</tr>
<tr>
<td></td>
<td> <math>=\,\,\lim_{h\rightarrow0}\frac{h}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
</tr>
<tr>
<td></td>
<td> <math>=\,\,\lim_{h\rightarrow0}\frac{1}{\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
<tr>
<td></td>
<td> <math>=\,\,\frac{1}{2\sqrt{x-5}}.</math></td>
</table>
|-
|Notice that this is the same result you would get using our more convenient "rules of integration", including the chain rule, but that's not the point of this problem. You specifically need to treat the derivative as a limit.
|}

[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 !Step 2: &nbsp;
 |-
-|We can't plug in <math style="vertical-align: 0%">h=0</math>, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:
+|We can't plug in <math style="vertical-align: 0%">h=0</math>, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:<br><br>
 <table>
    <tr>
@@ Line 49: / Line 50: @@
 </table>
 |-
-|Notice that this is the same result you would get using our more convenient "rules of integration", including the chain rule, but that's not the point of this problem. You specifically need to treat the derivative as a limit.
+|<br>Notice that this is the same result you would get using our more convenient "rules of integration," including the chain rule, but that's not the point of this problem. You specifically need to treat the derivative as a limit.
 |}

@@ Line 29: / Line 29: @@
 |We can't plug in <math style="vertical-align: 0%">h=0</math>, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:
 <table>
-<tr>
+  <tr>
-<td>&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}\cdot\frac{\sqrt{x+h-5}+\sqrt{x-5}}{\sqrt{x+h-5}+\sqrt{x-5}}</math></td>
+    <td>&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}\cdot\frac{\sqrt{x+h-5}+\sqrt{x-5}}{\sqrt{x+h-5}+\sqrt{x-5}}</math></td>
-<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{\left(\sqrt{x+h-5}\right)^{2}-\left(\sqrt{x-5}\right)^{2}}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
+    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{\left(\sqrt{x+h-5}\right)^{2}-\left(\sqrt{x-5}\right)^{2}}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
-</tr>
+  </tr>
-<tr>
+  <tr>
-<td></td>
+    <td></td>
-<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{x+h-5-\left(x-5\right)}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math> </td>
+    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{x+h-5-\left(x-5\right)}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math> </td>
-</tr>
+  </tr>
-<tr>
+  <tr>
-<td></td>
+    <td></td>
-<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{h}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
+    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{h}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
-</tr>
+  </tr>
-<tr>
+  <tr>
-<td></td>
+    <td></td>
-<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{1}{\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
+    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{1}{\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
-<tr>
+  <tr>
-<td></td>
+    <td></td>
-<td>&nbsp; <math>=\,\,\frac{1}{2\sqrt{x-5}}.</math></td>
+    <td>&nbsp; <math>=\,\,\frac{1}{2\sqrt{x-5}}.</math></td>
 </table>
 |-

← Older revision		Revision as of 18:36, 2 April 2015
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	\|}		\|}

		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!Final Answer:
		+	\|-
		+	\|Note that no points would be given for only a correct answer. This exam question is about correctly using the definition of a derivative. However, the correct answer is
		+	\|-
		+	\|     <math>f'(x)\,\,=\,\,\frac{1}{2\sqrt{x-5}}.</math>
		+	\|}
	[[022_Exam_1_Sample_A\|'''<u>Return to Sample Exam</u>''']]		[[022_Exam_1_Sample_A\|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}.</math>
 |-
-|The goal in solving this is to plug the values <math style="vertical-align: -10%">x+h</math> and <math style="vertical-align: 0%">x</math> into the appropriate <math style="vertical-align: -20%">f</math> in the numerator, and then find a way to cancel the <math style="vertical-align: 0%">h</math> in the denominator.  Unlike simplifying a  rational expression containing radicals, here it's appropriate to have a radical in the denominator.
+|The goal in solving this is to plug the values <math style="vertical-align: -8%">x+h</math> and <math style="vertical-align: 0%">x</math> into the appropriate <math style="vertical-align: -20%">f</math> in the numerator, and then find a way to cancel the <math style="vertical-align: 0%">h</math> in the denominator.  Unlike simplifying a  rational expression containing radicals, here it's appropriate to have a radical in the denominator.
 |-
 |Again, the goal is to cancel the <math style="vertical-align: 0%">h</math>.

@@ Line 22: / Line 22: @@
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(x)\,=\,\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\,=\,\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"