Science:Math Exam Resources/Courses/MATH104/December 2014/Question 05 (a)

MATH104 December 2014

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Question 05 (a)
Let $f(x)={\frac {e^{x}}{x^{2}}}$. Find the critical point of $f(x)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How are critical points and derivatives related?

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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. First, we determine $f'(x)$; rewrite $f(x)=e^{x}x^{-2}$ and use product rule on $h(x)=e^{x},g(x)=x^{-2}$: ${\begin{aligned}f'(x)&=h(x)g'(x)+h'(x)g(x)\\&=e^{x}x^{-2}-2x^{-3}e^{x}\\&={\frac {e^{x}}{x^{2}}}-{\frac {2e^{x}}{x^{3}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}\end{aligned}}$ The critical points of $f$ are at $x$ where $f'(x)=0$ and at values $x$ where $f'(x)$ does not exist. We observe that at $x=0$, $f'(x)$ does not exist and hence, there is a veritcal asymptote at $x=0$. Let $f'(x)=0$ and find $x$: $0={\frac {e^{x}(x-2)}{x^{3}}}$ Thus $e^{x}=0{\text{ or }}(x-2)=0$. Since $e^{x}>0$ for all $x\in \mathbb {R}$, we have that $x-2=0$ and hence $x=2$. So the critical point of $f$ occurs at $x=2$.

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