Answer
The force is attractive if the separation is larger than the equilibrium separation.
Work Step by Step
$U = \frac{A}{r^{12}}-\frac{B}{r^6}$
$\frac{dU}{dr} = \frac{6B}{r^7}-\frac{12A}{r^{13}}$
Suppose that $\frac{dU}{dr} \gt 0$:
$\frac{dU}{dr} = \frac{6B}{r^7}-\frac{12A}{r^{13}} \gt 0$
$\frac{6B}{r^7} \gt \frac{12A}{r^{13}}$
$r \gt \sqrt[6] {\frac{2A}{B}}$
Since $F = -\frac{dU}{dr}$, then $F \lt 0$ when $r \gt \sqrt[6] {\frac{2A}{B}}$
Therefore, the force is attractive if the separation is larger than the equilibrium separation.
