The potential energy of a 1 kg particle free to move along the x-axis is given by  V(x)  =    x44−x22  Joules.   The total Mechanical Energy of the particle 2J. Then the maximum speed is,

total energy TE=2 Jfor maximum speed⇒KE is maximum, ⇒PE is minimumPE=V(x)=x44-x22  differentiate with respect to x , to find minimum PEdVdx=4x34-2x2dVdx=x3-xfor maximum or miinimum dVdx=0x3-x=0 ⇒x2=1 ⇒x=±1for minimum value of PE consider x=-1V(x)=x44-x22substitute  x=-1V(x)=-14=PE minimumTE=KE+PEKEmax=TE-PEminKEmax=2--14KEmax=9412mv2=94v=32m/s