A particle of mass 0.50 kg moves along the positive x-axis…

Step 1. The force. $F = -\dfrac{dU}{dx} = \dfrac{2A}{x^3} - \dfrac{B}{x^2}$. It is zero at $x_0 = \dfrac{2A}{B} = 1.0\ \text{m}$, where $U(x_0) = 2.0\ \text{J} - 4.0\ \text{J} = -2.0\ \text{J}$.

Closer in ($x < x_0$), $F > 0$ pushes the particle outwards; further out, $F < 0$ pulls it back in. Either way the force points back to $x_0$, so the equilibrium is stable: $U$ has its lowest value there.

Step 2. Released at $0.60\ \text{m}$. The force at the start is $F = 18.52\ \text{N} - 11.11\ \text{N} = 7.41\ \text{N}$, away from the origin. The total energy is

The particle turns back where $U(x) = E$: $\dfrac{A}{x^2} - \dfrac{B}{x} = E$ gives $Ex^2 + Bx - A = 0$, that is $10x^2 - 36x + 18 = 0$ with $x$ in metres. The roots are $x = 0.6\ \text{m}$ (the start) and $x = 3.0\ \text{m}$. So it moves between $0.6\ \text{m}$ and $3.0\ \text{m}$.

It is fastest at $x_0$, where $U$ is least: $\tfrac12 mv^2 = E - U(x_0) = 0.889\ \text{J}$, so $v = \sqrt{\dfrac{2 \times 0.889\ \text{J}}{0.50\ \text{kg}}} = 1.89\ \text{m/s}$.

Step 3. Released at $0.50\ \text{m}$. Now $E = U(0.50\ \text{m}) = 8\ \text{J} - 8\ \text{J} = 0$. At $x_0$ the kinetic energy is $0 - (-2.0\ \text{J}) = 2.0\ \text{J}$, so the greatest speed is $\sqrt{\dfrac{2 \times 2.0\ \text{J}}{0.50\ \text{kg}}} = 2.83\ \text{m/s}$.

Far away $U$ rises towards zero but stays below it, so $U < E$ at every point beyond the start. There is no turning point: the particle goes out for ever, slowing down, with its speed tending to zero. It never comes back.