009A Sample Final 2, Problem 6 - Revision history

Kayla Murray at 17:18, 1 December 2017

2017-12-01T17:18:40Z

Kayla Murray at 20:31, 18 March 2017

2017-03-18T20:31:07Z

Kayla Murray at 23:45, 7 March 2017

2017-03-07T23:45:11Z

Kayla Murray at 21:56, 6 March 2017

2017-03-06T21:56:34Z

Kayla Murray at 21:48, 6 March 2017

2017-03-06T21:48:46Z

Kayla Murray at 01:55, 19 February 2017

2017-02-19T01:55:39Z

Kayla Murray at 02:28, 18 February 2017

2017-02-18T02:28:48Z

Kayla Murray: Created page with "Compute ::a) $\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}$ ::b)
2017-02-18T02:13:59Z

Created page with "Compute ::a) <math style="vertical-align: -14px">\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}</math> ::b) <math st..."

New page
Compute

::a) <math style="vertical-align: -14px">\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}</math>

::b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin^2x}{3x}</math>

::c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math>

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'''Solution:'''

'''(a)'''

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'''(b)'''

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'''(c)'''

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[[009A_Sample_Final_2|'''Return to Sample Exam''']]

← Older revision		Revision as of 17:18, 1 December 2017
Line 4:		Line 4:

	<span class="exam">on the interval  <math style="vertical-align: -5px">[0,2].</math>		<span class="exam">on the interval  <math style="vertical-align: -5px">[0,2].</math>
		+	<hr>
		+	[[009A Sample Final 2, Problem 6 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|'''1.''' 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>
−	\|-
−	\|'''2.''' 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.~~
−	\|}

		+	[[009A Sample Final 2, Problem 6 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~	+	[[009A_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]
−	~~!Step 1:  ~~
−	\|-
−	~~\|To find the absolute maximum and minimum of  <math style="vertical-align: -5px">f(x)</math>  on the interval  <math style="vertical-align: -5px">~~[~~0,2],</math>~~
−	\|-
−	~~\|we need to find the critical points of  <math style="vertical-align: -5px">f(x).</math>~~
−	\|-
−	~~\|Using the Quotient Rule, we have~~
−	\|-
−	\|
−	~~       <math>\begin{array}{rcl}~~
−	~~\displaystyle{f'(x)} & = & \displaystyle{\frac{(1+x)(1-x)'-(1-x)(1+x)'}{(1+x)^2}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{-2}{(1+x)^2}.}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|We notice that  <math style="vertical-align: -6px">f'(x)\ne 0</math>  for any  <math style="vertical-align: 0px">x.</math>~~
−	\|-
−	~~\|So, there are no critical points in the interval  <math style="vertical-align: -5px">~~[~~0,2].</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we have~~
−	\|-
−	~~\|       <math>f(0)=1,~f(2)=-\frac{1}{3}.</math>~~
−	\|-
−	~~\|Therefore, the absolute maximum value for  <math style="vertical-align: -5px">f(x)</math>  is  <math style="vertical-align: -1px">1</math>~~
−	\|-
−	~~\|and the absolute minimum value for  <math style="vertical-align: -5px">f(x)</math>  is  <math style="vertical-align: -15px">-\frac{1}{3}.</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	\|       The absolute maximum value for  <math style="vertical-align: -5px">f(x)</math>  is  <math style="vertical-align: -1px">1</math>  and the absolute minimum value for  <math style="vertical-align: -5px">f(x)</math>  is  <math style="vertical-align: -15px">-\frac{1}{3}.</math>
−	\|}
−	~~[[009A_Sample_Final_2~~\|'''<u>Return to Sample Exam</u>''']]

@@ Line 50: / Line 50: @@
 |Now, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(0)=1,~f(2)=\frac{-1}{3}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(0)=1,~f(2)=-\frac{1}{3}.</math>
 |-
 |Therefore, the absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1</math>
 |-
-|and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -15px">\frac{-1}{3}.</math>
+|and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -15px">-\frac{1}{3}.</math>
 |}
@@ Line 61: / Line 61: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;The absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -15px">\frac{-1}{3}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;The absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -15px">-\frac{1}{3}.</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 19: / Line 19: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 42: / Line 40: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!Final Answer: &nbsp;
 |-
+|
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::<math>f(x)=\frac{1-x}{1+x}</math>
-<span class="exam">on the interval <math>[0,2].</math>
+<span class="exam">on the interval &nbsp;<math style="vertical-align: -5px">[0,2].</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
-<span class="exam">Compute
+<span class="exam"> Find the absolute maximum and absolute minimum values of the function
-::<span class="exam">a) <math style="vertical-align: -14px">\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}</math>
+::<math>f(x)=\frac{1-x}{1+x}</math>
-::<span class="exam">b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin^2x}{3x}</math>
+<span class="exam">on the interval <math>[0,2].</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' 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>
 |-
+|
 |-
+|'''2.''' 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>
 |-
+|
 |}
@@ Line 23: / Line 25: @@
 !Step 1: &nbsp;
 |-
+|To find the absolute maximum and minimum of &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on the interval &nbsp;<math style="vertical-align: -5px">[0,2],</math>
 |-
+|
 |-
+|We notice that &nbsp;<math style="vertical-align: -6px">f'(x)\ne 0</math>&nbsp; for any &nbsp;<math style="vertical-align: 0px">x.</math>
 |-
+|So, there are no critical points in the interval &nbsp;<math style="vertical-align: -5px">[0,2].</math>
 |}
@@ Line 35: / Line 48: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |-
+|and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -15px">\frac{-1}{3}.</math>
 |}
@@ Line 44: / Line 61: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;The absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -15px">\frac{-1}{3}.</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]