how do you find the abs max of func x^3-12x on the interval 0

Question

how do you find the abs max of func x^3-12x on the interval 0<x<4

amistre64 · Answer

take the first and second derivatives....

amistre64 · Answer

y = 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 · Answer

y' = 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 · Answer

yes it does so i just plug in 0 and 4 into the second deritave to find the max?

anonymous · Answer

because the answer is 16 and i have no idea how that it

amistre64 · Answer

you 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 · Answer

y(0) = 0

y(4) = 16.  16 is obviously higher than 0

amistre64 · Answer

so that would be the maximium point within the interval....

anonymous · Answer

thanks haha i feel dumb now i knew that