Here, Follow normal procedure of finding maxima and minima and observing saddle point
f(x)=3x4-16x3+24x2+37
f'(x)=12x3 -48x2+48x
as we know that to find Critical point we have to equate it to zero;
f'(x)=0
now, we get roots i.e=0,2,2
now again f"(x)=36x2-96x+48
f''(0)=48 >0 i.e local minima at 0
f''(2)=0 i.e we need furthur investigation
now at x=2 we got f''(2)=0 ; so on x=2 we have to investigate furthur that means we have to find f "'()
f "'()=72x-96
f "'(2)=144-96 =48
i.e f "'(2) not equal to ZERO so it is a saddle point (or point of inflection)
therefor here only one local minima exist so the answer is "B"