## Fundamentals of Physics Extended (10th Edition)

$U_{\max }=U_{0}+12\times 2-3\times 2^{2}=39J$
Let's assume that our starting point is $x=0$ so the work done by conservative force at any given $x$ will be $W=\int ^{x}_{0}F\left( x\right) dx=\int ^{x}_{0}\left( 6x-12\right) =3x^{2}-12x(1)$ So the change in potential energy for conservative force will be: $\Delta U=-W\left( 2\right)$ So from (1) and (2) we get $\Delta U=-W=12x-3x^{2}(3)$ And also $\Delta U=U\left( x\right) -U_{0}\left( 4\right)$ From (3) and (4) we get: $U\left( x\right) =U_{0}+12x-3x^{2}$ in order $U(x)$ to be maximum positive $12x-3x^{2}$ must have its maximum value becouse $U_{0}$ is constant so firstly lets calculate maximum value of $12x-3x^{2}$ İn order to calculate $12x-3x^{2}=3\left( 4x-x^{2}\right) =-3\left( x^{2}-4x\right) =-3\left( x^{2}-4x+4\right) +12=12-3\left( x-2\right) ^{2}$ we see here it will be maximum when $x=2$ so maximum ptential energy will be $U_{\max }=U_{0}+12\times 2-3\times 2^{2}=39J$