Find the absolute extrema of the function on the closed interval. y = 3x2/3 − 2x, [−1, 1]

Y'=2x^-1/3-2=0, u get x^1/3=1, so x=1, then y=1, but interval [-1,1], we have 3 values of x:-1,0,1 
x=1 so y=1, x=0 so y=0, x=-1 so y=5 To get the extrema, derive the function. You get y' = 2x^-1/3 - 2. Set this equal to zero, and you get x=0 as the location of a critical point. Since you are on a closed interval [-1, 1], those points can also have an extrema

X=1  min at (0,0) max at (-1,5)