Equilibrium solutions of a nonlinear mass spring system. 
Consider the mass-spring system governed by the differential equation,
  $$ m\ddot{x}=-F(x) $$
Where $x(t)$ is the time-dependent position displacement of the mass.
  The spring force is a nonlinear function of $x$ and given by,
  $$F(x)=\alpha x-\beta x^3= \frac{dV}{dx}$$ 
  Where $\alpha$ and  $\beta$ are positive constants. 
Sketch the function $V(x)$ and show from this plot that the first equation has three stationary points. Can you judge from the plot of $V(x)$ which of these stationary points are stable and which are unstable? Assume that $V(0)=0$

Now by integrating $\frac{dV}{dx}$, I got 
$$V(x)=\frac{1}{2} \alpha x^2-\frac{1}{4}\beta x^4 $$
We see that $V(x)$ is the potential energy of the mass as a function of displacement.  Plotting this gave me something like this 

Where stationary points are at $x=0, x=\pm \sqrt{\frac{2\alpha}{\beta}}$. I got stuck here and according to the provided solution, the local minimality of $V(x)$ at $0$ implies stability and the maximality of $V(x)$ at $x=\pm \sqrt{\frac{2\alpha}{\beta}}$ implies instability. There is no further explanation and I am extremely confused. I am also not sure how showing $V(x)$ is stationary at the given points shows that $x(t)$ is also stationary.
I understand that if $\frac{dV}{dx}=0$ then $\ddot{x}=0$. But my understanding was that $\dot{x}$ also needs to be $0$ for $x(t)$ to be stationary. In fact for $V(x)$ where $x=0$ which is the local minima, $\dot{x}$ cannot be $0$ since If the potential energy is at its minimum, the kinetic energy has to be at its maximum value (conservation of energy) which means that $\dot{x}$ has to be the maximum value. So $\frac{dx}{dt}\neq 0$ which contradicts with my understanding that $\frac{dx}{dt}$ must equal $0$ at stationary points.
I would appreciate any help clarifying my confusion. 
Edit: Three dimensional phase portrait with $x$, $\dot{x}$ and $\ddot{x}$ as axis. 

 A: Taking
$$
m\dot x\ddot x + F(x)\dot x = 0\to \frac 12\frac{d}{dt}(m\dot x)^2+\frac{d}{dt}\left(\frac 12 \alpha x^2-\frac 14 \beta x^4\right) = 0
$$
with the integral
$$
m\dot x^2+\alpha x^2-\frac 12 \beta x^4 = C_0
$$
Equilibrium is attained when $\dot x = 0$ or
$$
\alpha x^2 - \frac 12 \beta x^4= C_0
$$
or at
$$
x^* = \pm \sqrt{\frac{\alpha \pm\sqrt{\alpha ^2-2 \beta  C_0}}{\beta }}
$$
The equilibrium qualification can be done depending on $\alpha^2-2\beta C_0$ and also on characterizing those feasible points as sources saddle points or sinks.
NOTE
If $C_0 = 0$ then
$$
x^* = \left\{-\sqrt{\frac{2\alpha}{\beta}},0,\sqrt{\frac{2\alpha}{\beta}}\right\}
$$
Attached a plot for $m = 1,\alpha = 1, \beta = 1$ showing the phase plane movement partial orbits, with $C_0 = 1$ (red), $C_0 = 0.5$ (blue) and $C_0 = 0.2$ (green)

Including viscous dissipation as $-k \dot x$ with $k = \frac 18$ we have the path from $\dot x(0) = 0.7, x(0) = 0$  (in red)

NOTE
Near $x = 0$ the linearized dynamic system behaves as
$$
m\ddot x + \alpha x = 0
$$
with solution
$$
x_0(t) = C_1 \sin \left(\frac{\sqrt{\alpha } t}{\sqrt{m}}\right)+C_2 \cos \left(\frac{\sqrt{\alpha } t}{\sqrt{m}}\right)
$$
and near $x = \pm\sqrt{\frac{2\alpha}{\beta}}$ the linearized movement is
$$
m\ddot x -2\alpha x = 0
$$
with solution
$$
x_{\sqrt{\frac{2\alpha}{\beta}}}(t) = C_1 e^{\frac{\sqrt{2\alpha } t}{\sqrt{m}}}+C_2 e^{-\frac{\sqrt{2\alpha } t}{\sqrt{m}}}\ \ \mbox{(unstable)}
$$
