Difference between revisions of "009A Sample Midterm 3, Problem 4"

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(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">(...")
 
 
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<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>
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<span class="exam"> Find the derivatives of the following functions. Do not simplify.
  
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<span class="exam">(a)&nbsp; <math style="vertical-align: -16px">f(x)=\frac{(3x-5)(-x^{-2}+4x)}{x^{\frac{4}{5}}}</math>
  
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<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>
!Foundations: &nbsp;
 
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|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
 
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|&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>
 
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<hr>
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[[009A Sample Midterm 3, Problem 4 Solution|'''<u>Solution</u>''']]
  
'''Solution:'''
 
  
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[[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
!Step 1: &nbsp;
 
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|First, we need to calculate the slope of the tangent line.
 
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|Let &nbsp;<math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
 
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|From Problem 3, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=-\frac{3}{\sqrt{-2x+5}}.</math>
 
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|Therefore, the slope of the tangent line is
 
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&nbsp; &nbsp; &nbsp; &nbsp; <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>
 
|}
 
  
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!Step 2: &nbsp;
 
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|Now, the tangent line has slope &nbsp;<math style="vertical-align: -1px">m=-1</math>
 
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|and passes through the point &nbsp;<math style="vertical-align: -5px">(-2,9).</math>
 
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|Hence, the equation of the tangent line is
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x+2)+9.</math>
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x+2)+9</math>
 
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|
 
|}
 
 
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 14:24, 9 November 2017

Find the derivatives of the following functions. Do not simplify.

(a) 

(b)    for  


Solution


Detailed Solution


Return to Sample Exam