Physics - W.E.P. Question with Solution | TestHub

For equilibrium, the force $F=-dU/dx$ must be zero.$$F=-\frac{d}{dx}(4x^3-12x^2+9x+10)=-(12x^2-24x+9)$$ Setting $F=0$: $12x^2-24x+9=0⟹4x^2-8x+3=0$. Factoring the quadratic equation: $(2x-1)(2x-3)=0$, which gives $x=1/2$ or $x=3/2$. For stable equilibrium, $d^{2}U/d{x}^2>0$.$$\frac{d^{2}U}{d{x}^2}=\frac{d}{dx}(12x^2-24x+9)=24x-24$$ At $x=1/2$, $d^{2}U/d{x}^2=24(1/2)-24=12-24=-12<0$ (unstable equilibrium). At $x=3/2$, $d^{2}U/d{x}^2=24(3/2)-24=36-24=12>0$ (stable equilibrium). Thus, the position of stable equilibrium is $x=3/2$.