009A Sample Midterm 1, Problem 3 - Revision history

Kayla Murray at 01:29, 6 November 2017

2017-11-06T01:29:10Z

Kayla Murray at 16:28, 5 November 2017

2017-11-05T16:28:05Z

Kayla Murray at 16:24, 5 November 2017

2017-11-05T16:24:26Z

Kayla Murray at 14:51, 3 November 2017

2017-11-03T14:51:58Z

Kayla Murray at 18:21, 18 March 2017

2017-03-18T18:21:23Z

Kayla Murray at 23:50, 26 February 2017

2017-02-26T23:50:25Z

Kayla Murray at 23:38, 26 February 2017

2017-02-26T23:38:05Z

Kayla Murray at 23:51, 18 February 2017

2017-02-18T23:51:54Z

Kayla Murray at 23:48, 18 February 2017

2017-02-18T23:48:08Z

Kayla Murray at 21:57, 18 February 2017

2017-02-18T21:57:56Z

@@ Line 16: / Line 16: @@
 <span class="exam">(d) Is &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; continuous at &nbsp;<math style="vertical-align: -1px">x=1?</math>&nbsp; Briefly explain.
 <hr>
 [[File:9ASM1P3.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 17: / Line 17: @@
 <hr>
-[[File:9ASM1P3.jpg|600px]]
+[[File:9ASM1P3.jpg|600px|thumb|center]]
 [[009A Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 17: / Line 17: @@
 <hr>
-(insert picture of handwritten solution)
+[[File:9ASM1P3.jpg|600px]]
 [[009A Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Let &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
+<span class="exam">Consider the following function &nbsp;<math style="vertical-align: -5px"> f:</math>
-<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
+<span class="exam">(a) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>
-<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1).</math>
+<span class="exam">(b) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<span class="exam">(d) Is &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; continuous at &nbsp;<math style="vertical-align: -1px">x=1?</math>&nbsp; Briefly explain.
-!Foundations: &nbsp;
+<hr>
-'''Solution:'''
+[[009A Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' &nbsp;<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
+|'''1.''' Recall
 |-
 |'''2.''' 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

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' '''Limit Definition of Derivative'''
+|'''1.''' &nbsp;<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: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
+|'''2.''' 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
 |-
-|'''2.''' '''Equation of a tangent line'''
+|&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 27: / Line 23: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
 |-
 |Using the limit definition of the derivative, we have
@@ Line 67: / Line 63: @@
 !Step 1: &nbsp;
 |-
-|We start by finding the slope of the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math>
+|We start by finding the slope of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1).</math>
 |-
 |Using the derivative calculated in part (a), the slope is
@@ Line 83: / Line 79: @@
 !Step 2: &nbsp;
 |-
-|Now, the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1)</math>
+|Now, the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1)</math>
 |-
-|has slope <math style="vertical-align: -13px">m=\frac{3}{2}</math> and passes through the point <math style="vertical-align: -5px">(2,1).</math>
+|has slope &nbsp;<math style="vertical-align: -13px">m=\frac{3}{2}</math>&nbsp; and passes through the point &nbsp;<math style="vertical-align: -5px">(2,1).</math>
 |-
 |Hence, the equation of this line is

@@ Line 15: / Line 15: @@
 |'''2.''' '''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 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
-|Let <math>f(x)=\sqrt{3x-5}.</math>
+|Let <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
 |-
 |Using the limit definition of the derivative, we have
@@ Line 67: / Line 67: @@
 !Step 1: &nbsp;
 |-
-|We start by finding the slope of the tangent line to <math>f(x)=\sqrt{3x-5}</math> at <math>(2,1).</math>
+|We start by finding the slope of the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math>
 |-
 |Using the derivative calculated in part (a), the slope is
@@ Line 83: / Line 83: @@
 !Step 2: &nbsp;
 |-
-|Now, the tangent line to <math>f(x)=\sqrt{3x-5}</math> at <math>(2,1)</math>
+|Now, the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1)</math>
 |-
-|has slope <math>m=\frac{3}{2}</math> and passes through the point <math>(2,1).</math>
+|has slope <math style="vertical-align: -13px">m=\frac{3}{2}</math> and passes through the point <math style="vertical-align: -5px">(2,1).</math>
 |-
 |Hence, the equation of this line is

@@ Line 1: / Line 1: @@
-<span class="exam">Let <math>y=\sqrt{3x-5}.</math>
+<span class="exam">Let <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
-<span class="exam">(a) Use the definition of the derivative to compute <math>\frac{dy}{dx}</math> for <math>y=\sqrt{3x-5}.</math>
+<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
-<span class="exam">(b) Find the equation of the tangent line to <math>y=\sqrt{3x-5}</math> at <math>(2,1).</math>
+<span class="exam">(b) Find the equation of the tangent line to <math style="vertical-align: -5px">y=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math>