graphing $\frac{x^3-x+1}{x^2}$ I want to graph:
$$f(x) = \frac{x^3-x+1}{x^2}$$
so I took the first derivative:
$$f'(x) = \frac{x^3+x-2}{x^3}$$
but this function is hard to find the signals. In other words, it's hard to find where the function is increasing or decreasing. Also, the second derivative is even worse. I'm assuming it shouldn't be all that hard to find the signals, because it's as exercise on a book.
Any ideas? Do I really have to find the roots of the numerator?
 A: Study of sign of $f':$
Write the numerator as
$$x^3+x-2=(x-1)(x^2+x+2).$$ Now, $(x-1)(x^2+x+2)=0\iff x=1.$ So, 
$$\begin{array}{ccc} x & (-\infty,1) & (1,\infty) \\ \mathrm{sign}(x^3+x-2) & - & +\end{array}$$
It is clear that $x^3=0\iff x=0.$ Now
$$\begin{array}{ccc} x & (-\infty,0) & (0,\infty) \\ \mathrm{sign}(x^3) & - & +\end{array}$$
Thus
$$\begin{array}{cccc} x & (-\infty,0)& (0,1) & (1,\infty) \\ \mathrm{sign}(f') & + & - & +\end{array}$$
So, $f$ is increasing in $(-\infty,0),$ decreasing in $(0,1)$ and increasing in $(1,\infty).$ Since it changes from decreasing to increasing at $x=1$ it has a local minimum at $x=1.$ (Note that $0$ doesn't belong to the domain.)
Note that the study of $$f''(x)=\frac{2(3-x)}{x^4}$$ is simpler. Repeating the same process:
$$2(3-x)=0\iff x=3$$ and
$$\begin{array}{ccc} x & (-\infty,3) & (3,\infty) \\ \mathrm{sign}(2(3-x)) & + & -\end{array}$$
$$x^4=0\iff x=0$$ and
$$\begin{array}{ccc} x & (-\infty,0) & (0,\infty) \\ \mathrm{sign}(x^4) & + & -\end{array}$$ Finally
$$\begin{array}{cccc} x & (-\infty,0) & (0,3) & (3,\infty) \\ \mathrm{sign}(f'') & + & + & -\end{array}$$ Thus $f$ is convex on $(-\infty,0)$ and $(0,3)$ and concave on $(3,\infty).$ Since it changes curvature at $x=3$ $f$ has an inflection point at $x=3.$
