<s>[INST] How to find maximum/minimum of $y=\frac{x(x^2-x+2)}{x^2-9}$? How can I find minimum and maximum of $y=\frac{x(x^2-x+2)}{x^2-9}$?
In other words, points, where $y'=0$.
My current steps are:
$$y=\frac{x(x^2-x+2)}{x^2-9}=x-1+\frac{11x-9}{x^2-9}$$
$$\frac{dy}{dx}=1+\frac{11(x^2-9)-2x(11x-9)}{(x^2-9)^2}$$
\begin{align}
y'=0 &\Rightarrow 1+\frac{11x^2-99-22x^2+18x}{(x^2-9)^2}=0 \\
&\Rightarrow\frac{11x^2-99-22x^2+18x+(x^2-9)^2}{(x^2-9)^2}=0 \\ 
&\Rightarrow 11x^2-99-22x^2+18x+x^4-18x^2+81=0 \\
&\Rightarrow x^4-29x^2+18x-18=0
\end{align}
Mathematica 8 gives me two complex and two real roots, but I have no idea how to solve 4th order polynomial equations, nor I am sure if I really have to do this, considering that nothing this ''complex'' should be done in second semester calculus homework. 
 [/INST] Consider $$f(x)=\frac{x(x^2-x+2)}{x^2-9}$$
as $x \to \pm \infty$. Note the $x^2$ terms from the numerator and denominator cancel out, so you are left with
$$
f(x) = \frac{x(x^2-x+2)}{x^2-9} \approx x \text{ for very large } x.
$$
As a result,
$$\lim_{x \to \pm \infty} f(x) = \pm \infty$$
and it is unbounded on both ends.
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