009A Sample Midterm 3, Problem 4 - Revision history

Kayla Murray at 01:57, 7 November 2017

2017-11-07T01:57:28Z

Kayla Murray at 17:50, 3 November 2017

2017-11-03T17:50:56Z

Kayla Murray at 17:21, 13 March 2017

2017-03-13T17:21:01Z

Kayla Murray at 00:45, 27 February 2017

2017-02-27T00:45:13Z

Kayla Murray at 22:51, 18 February 2017

2017-02-18T22:51:06Z

Kayla Murray at 22:47, 18 February 2017

2017-02-18T22:47:04Z

Kayla Murray at 19:54, 18 February 2017

2017-02-18T19:54:04Z

Kayla Murray at 23:26, 17 February 2017

2017-02-17T23:26:49Z

Kayla Murray at 23:08, 17 February 2017

2017-02-17T23:08:15Z

Kayla Murray at 21:12, 16 February 2017

2017-02-16T21:12:04Z

@@ Line 7: / Line 7: @@
 <hr>
-(insert picture of handwritten solution)
+[[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-[[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[File:9ASM3P4.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Find the equation of the tangent line to &nbsp;<math style="vertical-align: -4px">y=3\sqrt{-2x+5}</math>&nbsp; at &nbsp;<math style="vertical-align: -4px">(-2,9).</math>
+<span class="exam"> Find the derivatives of the following functions. Do not simplify.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<span class="exam">(b)&nbsp; <math>g(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{\pi}</math>&nbsp; for &nbsp;<math style="vertical-align: 0px">x>0.</math>
-'''Solution:'''
+(insert picture of handwritten solution)
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 24: / Line 24: @@
 |From Problem 3, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{-3}{\sqrt{-2x+5}}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=-\frac{3}{\sqrt{-2x+5}}.</math>
 |-
 |Therefore, the slope of the tangent line is
@@ Line 32: / Line 32: @@
 \displaystyle{m} & = & \displaystyle{f'(-2)}\\
 &&\\
-& = & \displaystyle{\frac{-3}{\sqrt{-2(-2)+5}}}\\
+& = & \displaystyle{-\frac{3}{\sqrt{-2(-2)+5}}}\\
 &&\\
-& = & \displaystyle{\frac{-3}{\sqrt{9}}}\\
+& = & \displaystyle{-\frac{3}{\sqrt{9}}}\\
 &&\\
 & = & \displaystyle{-1.}

@@ Line 1: / Line 1: @@
-<span class="exam">Find the equation of the tangent line to <math style="vertical-align: -4px">y=3\sqrt{-2x+5}</math> at <math style="vertical-align: -4px">(-2,9).</math>
+<span class="exam">Find the equation of the tangent line to &nbsp;<math style="vertical-align: -4px">y=3\sqrt{-2x+5}</math>&nbsp; at &nbsp;<math style="vertical-align: -4px">(-2,9).</math>
@@ Line 5: / Line 5: @@
 !Foundations: &nbsp;
 |-
-|'''Equation of a tangent line'''
+|The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; The equation of the tangent line to <math style="vertical-align: -5px">f(x)</math> at the point <math style="vertical-align: -5px">(a,b)</math> is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
 |}
@@ Line 20: / Line 20: @@
 |First, we need to calculate the slope of the tangent line.
 |-
-|Let <math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
 |-
 |From Problem 3, we have
@@ Line 43: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|Now, the tangent line has slope <math style="vertical-align: -1px">m=-1</math>
+|Now, the tangent line has slope &nbsp;<math style="vertical-align: -1px">m=-1</math>
 |-
-|and passes through the point <math style="vertical-align: -5px">(-2,9).</math>
+|and passes through the point &nbsp;<math style="vertical-align: -5px">(-2,9).</math>
 |-
 |Hence, the equation of the tangent line is

@@ Line 7: / Line 7: @@
 |'''Equation of a tangent line'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; The equation of the tangent line to <math>f(x)</math> at the point <math>(a,b)</math> is
+|&nbsp; &nbsp; &nbsp; &nbsp; The equation of the tangent line to <math style="vertical-align: -5px">f(x)</math> at the point <math style="vertical-align: -5px">(a,b)</math> is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=m(x-a)+b</math> where <math>m=f'(a).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
 |}
@@ Line 20: / Line 20: @@
 |First, we need to calculate the slope of the tangent line.
 |-
-|Let <math>f(x)=3\sqrt{-2x+5}.</math>
+|Let <math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
 |-
 |From Problem 3, we have
@@ Line 43: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|Now, the tangent line has slope <math>m=-1</math>
+|Now, the tangent line has slope <math style="vertical-align: -1px">m=-1</math>
 |-
-|and passes through the point <math>(-2,9).</math>
+|and passes through the point <math style="vertical-align: -5px">(-2,9).</math>
 |-
 |Hence, the equation of the tangent line is

← Older revision		Revision as of 22:47, 18 February 2017
Line 1:		Line 1:
−	<span class="exam">Find the equation of the tangent line to <math>y=3\sqrt{-2x+5}</math> at <math>(-2,9).</math>	+	<span class="exam">Find the equation of the tangent line to <math style="vertical-align: -4px">y=3\sqrt{-2x+5}</math> at <math style="vertical-align: -4px">(-2,9).</math>

@@ Line 14: / Line 14: @@
 !Step 1: &nbsp;
 |-
+|First, we need to calculate the slope of the tangent line.
 |-
+|
 |}
@@ Line 22: / Line 39: @@
 !Step 2: &nbsp;
 |-
+|Now, the tangent line has slope <math>m=-1</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x+2)+9.</math>
 |}
@@ Line 31: / Line 52: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x+2)+9</math>
 |-
+|
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]