Science:Math Exam Resources/Courses/MATH110/April 2018/Question 03 (b)

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Question 03 (b)
For what value of the constant ${\textstyle a}$ does the graph of ${\textstyle y=xe^{-ax}}$ have a horizontal tangent line at ${\textstyle x=2}$ ?

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Hint
The graph of a function ${\textstyle f(x)}$ has a horizontal tangent line at ${\textstyle x=2}$ if ${\textstyle f'(2)=0}$ .

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let ${\textstyle f(x)=xe^{-ax}}$ . Then the graph of ${\textstyle y}$ has a horizontal tangent line at ${\textstyle x=2}$ if ${\textstyle f'(2)=0}$ . The first step is to calculate ${\textstyle f'(2)}$ with the indeterminate constant ${\textstyle a}$ . Using the product rule, we get $y'=e^{-ax}+x\cdot (-a)\cdot e^{-ax}=(1-ax)e^{-ax}.$ So at ${\textstyle x=2}$ , we have ${\textstyle f'(2)=(1-2a)e^{-2a}}$ . Now we have to solve the equation ${\textstyle (1-2a)e^{-2a}=0}$ for ${\textstyle a}$ . Noting that ${\textstyle e^{-2a}}$ is never zero (in fact, it is always positive), we can divide both sides by ${\textstyle e^{-2a}}$ , which leaves ${\textstyle 1-2a=0}$ . It follows that ${\textstyle a=1/2}$ . Answer: The graph of ${\textstyle y=xe^{-ax}}$ has a horizontal tangent line at ${\textstyle x=2}$ if and only if ${\textstyle {\color {blue}a=1/2}}$ .