University Physics with Modern Physics (14th Edition)

$F = -\frac{dU}{dx}$ $F = -4\alpha ~x^3$ $F = (-4)(0.630~J/m^4)~x^3$ $F = (-2.52~J/m^4)~x^3$ We can find the force when the particle is at x = -0.800 m. $F = (-2.52~J/m^4)~x^3$ $F = (-2.52~J/m^4)(-0.800~m)^3$ $F = 1.29~N$ The force is 1.29 N and it is directed toward the +x-axis.