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. 
 A: 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.
A: You can use partial fractions here $$\frac{11x-9}{x^2-9}=\frac A{x+3}+\frac B{x-3}=\frac 7{x+3}+\frac 4{x-3}$$ so that $$y=x-1+\frac 7{x+3}+\frac 4{x-3}$$ and $$y'=1-\frac 7{(x+3)^2}-\frac 4{(x-3)^2}$$
Then you should be able to sketch the functions to see what is happening, and to see that any real zeros of the derivative will be at local extrema.
Once your sketch shows you what is going on, you should be able to find suitable values of $x$ on either side of the zeros of $y'$ - the intermediate value theorem then tells you zeros exist (stay away from $x=\pm 3$), and you can use standard numerical methods to estimate the roots.
Splitting the function into simpler components greatly simplifies sketching, and sketching greatly helps if the situation looks a bit complicated.
