009A Sample Midterm 3, Problem 4 - Revision history

MathAdmin at 22:24, 9 November 2017

2017-11-09T22:24:57Z

MathAdmin at 16:12, 7 November 2017

2017-11-07T16:12:44Z

MathAdmin: Created page with "Find the equation of the tangent line to $y=3\sqrt{-2x+5}$ at $(..."$

2017-04-09T18:56:32Z

Created page with "<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">(..."

New page

<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>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|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:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to calculate the slope of the tangent line.
|-
|Let  <math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
|-
|From Problem 3, we have
|-
|        <math>f'(x)=-\frac{3}{\sqrt{-2x+5}}.</math>
|-
|Therefore, the slope of the tangent line is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{f'(-2)}\\
&&\\
& = & \displaystyle{-\frac{3}{\sqrt{-2(-2)+5}}}\\
&&\\
& = & \displaystyle{-\frac{3}{\sqrt{9}}}\\
&&\\
& = & \displaystyle{-1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, the tangent line has slope  <math style="vertical-align: -1px">m=-1</math>
|-
|and passes through the point  <math style="vertical-align: -5px">(-2,9).</math>
|-
|Hence, the equation of the tangent line is
|-
|        <math>y=-1(x+2)+9.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        <math>y=-1(x+2)+9</math>
|-
|
|}
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

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 <hr>
-[[File:9ASM3P4.jpg|600px|thumb|center]]
+[[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[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:'''
+[[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[File:9ASM3P4.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

009A Sample Midterm 3, Problem 4 - Revision history

MathAdmin at 22:24, 9 November 2017

MathAdmin at 16:12, 7 November 2017

MathAdmin: Created page with "Find the equation of the tangent line to y=3\sqrt{-2x+5} at (..."

MathAdmin: Created page with "Find the equation of the tangent line to $y=3\sqrt{-2x+5}$ at $(..."$