check end points
f(-2) = -2+8 = 6
f(2) = 2-8 = -6
now find where derivative is zero
f' = 1 -3x^2
= 0 when x = 1/sqrt3 or x = -1/sqrt3
if x = 1/sqrt 3
f = 1/sqrt3 -1/3sqrt3 =(2/3sqrt3)
if x = -1/sqrt3
f = -1/sqrt3 +1/3sqrt 3
= -(2/3sqrt3)
so the max and min are at the end points max at -2 and min at +2
find the maximum and minimum value of the function f(x)= x-x^3 on the interval [-2,2]
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