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anonymous
 5 years ago
how do you find the abs max of func x^312x on the interval 0<x<4
anonymous
 5 years ago
how do you find the abs max of func x^312x on the interval 0<x<4

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0take the first and second derivatives....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = x^3  12x y' = 3x^2  12 y' = 0 will give critical points; y'' = 6x ; plug those points into y'' to see if they are (+) or () to find max and min

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y' = 3x^2  12x = 0 x(3x  12) = 0 when x = 0 or x = 4 are the critical points of your equation... but I have to wonder, does your interval INCLUDE 0 and 4? because the way you have it written it means everything between 0 and 4 EXCEPT for 0 and 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it does so i just plug in 0 and 4 into the second deritave to find the max?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the answer is 16 and i have no idea how that it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you plug it into the second derivative see how it "acts"; and according to its behaviour, there are a few possibilities of what it could be... But first, plug in 0 and 4 into your original equation to see what value they give..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y(0) = 0 y(4) = 16. 16 is obviously higher than 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so that would be the maximium point within the interval....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks haha i feel dumb now i knew that
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