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find the relative maximum and minimum of the equation 2x^3-3x^2-36x+2?
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take the derivative: y'=6x^2-6x-36
set it equal to 0 since it's defined everywhere
x^2-x-6=0
solve for x
then take the second derivative.. and plug in the x values that u get to see if it's a max or a min
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if second derivative is positive, then ur x value is a min, if negative, it's a max
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