The correct option is B x=3 and x=5 are points of equilibrium.
Given,
E=x33−4x2+15x−6
For equilibrium,
dEdx=0
Therefore, we get,
dEdx=ddx(x33−4x2+15x−6)=0⇒3x23−8x+15=0⇒x2−8x+15=0⇒(x−3)(x−5)=0⇒x=3 or x=5
Hence the body will be in equilibrium at x=3 and x=5.
If d2Edx2>0, then it is a stable equilibrium, If d2Edx2<0, then it is an unstable equilibrium.
d2Edx2=2x−8
At x=3, d2Edx2=2×3−8=−2
The system is in an unstable equilibrium at x=3
At x=5,
d2Edx2=2×5−8=2
The system is in a stable equilibrium at x=5.