A force parallel to the x-axis?

Question

A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces a potential energy U(x) given by U(x)= (1.14J/m^4)x^4.

What is the force (magnitude and direction) when the particle is at position x= -0.570m?

Answer 1

F = -dU/dx

F = -(1.14)4x³
F = 0.84448 N

F is approximately 0.844 N

Answer 2

the force is the negative of the derivative of U(x), so
F(x) = -1.14(4x^3) = -4.56x^3
F(-.570) = .844 N in the positive direction (I suppose to the right).

Answer 3

F = -dU/dx = -4.56 x^3 = +0.844 N