Science:Math Exam Resources/Courses/MATH104/December 2014/Question 05 (a)/Solution 1

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 $x=2$.