Autonomous equilibrium points Given $$\frac{dx}{dt}= 3x-x^2$$ 
I don't understand how $$x=0$$ is not semistable.
I get the following 0 points: $$x = 0, x = 3$$
Here are the values of $\frac{dx}{dt}$ I get when plugging in and my reasoning:
$-3 \to -18$
$-2 \to -10$
$-1 \to -4$
It would seem to me that clearly the slope gets closer to zero as we go up and would have the curve going up and arcing to the right and then flattening out but no the book has the curve going the opposite direction.....doesnt make any sense at all
$1 \to 2$ 
$1.5 \to 2.25$
$2 \to 2$
The slope increases then levels out and then goes down somehow giving an S shape between $x = 0$ and $x = 3$
Now for 
$6 \to -18$
$5 \to -10$
$4 \to  -4$
The slope gets nearer to zero as we near 3 so we get a kind of semi c shape going down and leveling out at $x = 3$ This turned out to be right. So I don't understand how my reasoning worked out in one instance but the same exact approach didn't work in the first part.

The zeroes--critical points--of the equation
$\dot x = 3x - x^2 \tag 1$
occur where
$\dot x = 0, \tag 2$
that is, where
$3x - x^2 = 0; \tag 3$
it is easy to see that the values of $x$ satisfying this quadratic are
$x = 0, 3; \tag 4$
the stability of these critical points is, in accord with the well-known theory, determined by the values of
$\dfrac{d(\dot x)}{dx} = \dfrac{d(3x - x^2)}{dx} = 3 - 2x \tag 5$
at these values of $x$; we have
$\dfrac{d(\dot x)}{dx}(0) = 3 > 0, \tag 6$
hence $0$ is unstable; and since
$\dfrac{d(\dot x)}{dx}(3) = -3 < 0, \tag 7$
$x = 3$ is a stable equilibrium of (1).
These calculations are simple and could mos' likely be performed in a few minutes during an exam; quicker, I'll warrant, that the amount of arithmetic/sketching a graphical solution requires; also, more rigorous.  Nevertheless, I thank my colleagues Quanto and Moo for their explanitory artistry.
