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| − | <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> | + | <span class="exam"> Find the derivatives of the following functions. Do not simplify. |
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| | + | <span class="exam">(a) <math style="vertical-align: -16px">f(x)=\frac{(3x-5)(-x^{-2}+4x)}{x^{\frac{4}{5}}}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(b) <math>g(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{\pi}</math> for <math style="vertical-align: 0px">x>0.</math> |
| − | !Foundations:
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| − | |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
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| − | | <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
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| | + | <hr> |
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| − | '''Solution:''' | + | [[009A Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[File:9ASM3P4.jpg|600px|thumb|center]] |
| − | !Step 1:
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| − | |First, we need to calculate the slope of the tangent line.
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| − | |Let <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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| − | | <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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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{m} & = & \displaystyle{f'(-2)}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{3}{\sqrt{-2(-2)+5}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{3}{\sqrt{9}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-1.}
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| − | \end{array}</math>
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| − | |} | |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, the tangent line has slope <math style="vertical-align: -1px">m=-1</math>
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| − | |and passes through the point <math style="vertical-align: -5px">(-2,9).</math>
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| − | |Hence, the equation of the tangent line is
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| − | | <math>y=-1(x+2)+9.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>y=-1(x+2)+9</math>
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| − | |}
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| | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
Find the derivatives of the following functions. Do not simplify.
(a) 
(b) 
for 
Detailed Solution with Background Information
Return to Sample Exam
