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find the maximum and minimum value of y=3x^3 - 4x +2
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\[\frac{ d (3x^3 - 4x +2)}{ dx} = 9x^2-4\]
9x^2-4=0 x= 2/3 x=-2/3 nw wat to do
as in the range of the function? its a cubic; odd integer degrees rise towards infinity and dive towards negative infinity
if your trying to determine the bumps and dips; take the derivative a second time, insert the zeros from that, and see if the signs are positive or negative
positive opens up; U which translates as a min negative opens Down n, which translates as a max
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