f′′(x)=36x2−96x+48
at x=0,f′′(x)=48>0 it means that x=0 is local minima.
but at x=2,f′′(x)=0 so we can't apply second derivative test.
So, we can apply third derivative test
3rd derivative = 72x- 96
At x=2, 3rd derivative >0 therefore local maxima? But ans do have 2 as local maxima.
What i am doing wrong?