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

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(Created page with "<span class="exam"> Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math>   for  <math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</mat...")
 
 
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<span class="exam"> Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for &nbsp;<math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</math>
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<span class="exam"> Let &nbsp;<math style="vertical-align: -3px">y=3\sqrt{2x+5},x\ge 0.</math>
  
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<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math style="vertical-align: -13px">\frac{dy}{dx}.</math>
  
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<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -3px">y=3\sqrt{2x+5}</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">(2,9).</math>
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!Foundations: &nbsp;
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<hr>
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|Recall
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
 
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[[009A Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']]
  
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'''Solution:'''
 
  
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[[009A Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
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!Step 1: &nbsp;
 
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|Let &nbsp;<math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
 
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|Using the limit definition of the derivative, we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2(x+h)+5}-3\sqrt{-2x+5}}{h}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2x+-2h+5}-3\sqrt{-2x+5}}{h}}\\
 
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&&\\
 
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& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{\sqrt{-2x+-2h+5}-\sqrt{-2x+5}}{h}.}
 
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\end{array}</math>
 
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!Step 2: &nbsp;
 
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|Now, we multiply the numerator and denominator by the conjugate of the numerator.
 
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|Hence, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f'(x)} & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(\sqrt{-2x+-2h+5}-\sqrt{-2x+5})}{h} \frac{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
 
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&&\\
 
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& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(-2x+-2h+5)-(-2x+5)}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
 
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&&\\
 
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& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2h}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
 
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&&\\
 
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& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2}{\sqrt{-2x+-2h+5}+\sqrt{-2x+5}}}\\
 
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& = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\
 
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& = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.}
 
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\end{array}</math>
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>-\frac{3}{\sqrt{-2x+5}}</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 12:22, 11 November 2017

Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=3\sqrt{2x+5},x\ge 0.}

(a) Use the definition of the derivative to compute   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}.}

(b) Find the equation of the tangent line to  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=3\sqrt{2x+5}}   at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2,9).}

Solution


Detailed Solution


Return to Sample Exam

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