009A Sample Midterm 1, Problem 3 - Revision history

MathAdmin at 21:41, 8 November 2017

2017-11-08T21:41:15Z

MathAdmin at 15:29, 7 November 2017

2017-11-07T15:29:24Z

MathAdmin at 02:53, 14 April 2017

2017-04-14T02:53:14Z

MathAdmin: Created page with "Let $y=\sqrt{3x-5}.$ (a) Use the definition of the derivative to compute $\fra..."$

2017-04-09T18:34:50Z

Created page with "Let <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math> (a) Use the definition of the derivative to compute <math>\fra..."

New page

Let  <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>

(a) Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math>   for  <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>

(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>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' Recall
|-
|       <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  <math style="vertical-align: -5px">f(x)</math>  at the point  <math style="vertical-align: -5px">(a,b)</math>  is
|-
|        <math style="vertical-align: -5px">y=m(x-a)+b</math>  where  <math style="vertical-align: -5px">m=f'(a).</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
|-
|Using the limit definition of the derivative, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3(x+h)-5}-\sqrt{3x-5}}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3x+3h-5}-\sqrt{3x-5}}{h}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we multiply the numerator and denominator by the conjugate of the numerator.
|-
|Hence, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\sqrt{3x+3h-5}-\sqrt{3x-5})}{h} \frac{(\sqrt{3x+3h-5}+\sqrt{3x-5})}{(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{(3x+3h-5)-(3x-5)}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3h}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3}{\sqrt{3x+3h-5}+\sqrt{3x-5}}}\\
&&\\
& = & \displaystyle{\frac{3}{\sqrt{3x-5}+\sqrt{3x-5}}}\\
&&\\
& = & \displaystyle{\frac{3}{2\sqrt{3x-5}}.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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
|-
|        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{f'(2)}\\
&&\\
& = & \displaystyle{\frac{3}{2\sqrt{3(2)-5}}}\\
&&\\
& = & \displaystyle{\frac{3}{2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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 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
|-
|        <math>y=\frac{3}{2}(x-2)+1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\frac{3}{2\sqrt{3x-5}}</math>
|-
|    '''(b)'''     <math>y=\frac{3}{2}(x-2)+1</math>
|}
[[009A_Sample_Midterm_1|'''Return to Sample Exam''']]

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

@@ Line 94: / Line 94: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{3}{2\sqrt{3x-5}}</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{3}{2\sqrt{3x-5}}</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>y=\frac{3}{2}(x-2)+1</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

009A Sample Midterm 1, Problem 3 - Revision history

MathAdmin at 21:41, 8 November 2017

MathAdmin at 15:29, 7 November 2017

MathAdmin at 02:53, 14 April 2017

MathAdmin: Created page with "Let y=\sqrt{3x-5}. (a) Use the definition of the derivative to compute \fra..."

MathAdmin: Created page with "Let $y=\sqrt{3x-5}.$ (a) Use the definition of the derivative to compute $\fra..."$