Hi darren86!

The total energy of your system is the kinetic energy plus the potential energy, which must be constant.

With a mass m moving along an x-axis, it takes the form:

E = (1/2)m xdot^2 + V(x) = constant

If you take the derivative with respect to t, you get:

m xdot xdotdot + V'(x) xdot = 0

which simplifies to:

m xdotdot = -V'(x)

You have already deduced which form V(x) takes.

To be a little more in line with physics, it is actually:

V(x) = m(ax^2/2-x^4/4) + V_{0}

where V

_{0} some constant, or more specifically the potential energy when x=0.

The stationary points occur when xdot = 0.

When this happens we have:

E = (1/2)m xdot^2 + V(x) = 0 + V(x)

In other words:

V(x) = m(ax^2/2-x^4/4) + V_{0} = E

Solve for x and you find your stationary points.

Can you analyze the graph of V(x) versus x?

And perhaps draw it?

You should find the V(x) takes certain maxima somewhere.

These are interesting points.

Can you find them?

In particular you'll have 2 maxima that have the same potential energy.

So if we release a test mass at one of these maxima with initial speed zero, it will "just" remain between these maxima.

And moreover the total energy E of the system will be equal to the potential energy in these maxima.

These maxima are what you are asked for.