f′(x)=5x4−20x3+15x2=5x2(x−1)(x−3). Critical points: x=0,1,3.
f′′(x)=10x(2x2−6x+3). f′′(0)=0 (inflexion), f′′(1)<0 (max), f′′(3)>0 (min).
f(1)=1−5+5−1=0 (local max value).
f(3)=243−405+135−1=−28 (local min value).
So (a), (b), (c), (d) are correct; (e) is wrong since inflexion is at x=0.