How do you find the maximum value of #3x^2-x^3#?

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Answer 1 · 2016-08-11T09:24:51

#4#

#y = 3x^2 - x^3 #

#y' = 6x - 3x^2 = 3x(2 - x)#

#y' = 0 implies x = 0, 2#

#y'' = 6 - 6x = 6(1-x)#

#y''(0) = 6 > 0# a min

#y''(2) = -6 < 0# ie a max

#y(2) = 4#, the max value of y

graph{3x^2 - x^3 [-12.66, 12.66, -6.33, 6.33]}