f'(x)=5x4-4x3=0 at extrema.
x3(5x-4)=0⇒x=0 and ⅘.
f"(x)=20x3-12x2=0 when x=0 and is positive when x= ⅘, indicating a minimum.
So there is a maximum or point of inflection at (0,0). x=0 is between -1 and 1 so it qualifies as a a maximum in the interval. However, we must also consider f(1)=0 and f(-1)=-2. The former is also zero and we know that there is a minimum at x=⅘, close to x=1, and f(⅘)<0 so 0 is also the absolute maximum.