Recent content by NyteBlayde

agreed, that's why i figured when i found the error while writing it out i figured id tell you :)

lol round 2 was short lived :P same damn answer lmao

damn straight :D

I know, i just did both extrema, i only needed min and there was an extraneous value, just wanted to do the math

lmao i feel so retarded, i pluged it into the area function instead of the original >.< good thing you caught me thx again :)

Alright, once again thanks for all your help lol, couldn't have done it without you :D. Now i can finalize these in word and put them away :)

yuck i hate dealing with radicals lol ok, i think i got it. if I am right, the extremes of the area are at P(.sqrt3 / 3 , .sqrt3 / 9) and Q( -.sqrt3 / 3 , -2.sqrt3 / 9 )

lol this going to be kinda ugly, +/- 1/.sqrt 3 (without rationalizing)

REPOST: wasnt sure if ud see it no since this is page 3 lol thanks a ton for helping out on that one :D back on the topic of the first question, u set 1/2(1-3x^2) = 0 so i should end up with 1/2 - 3/2x^2 = 0 and all i need to do is solve for x and then find my min right?

3 down 1 togo :D, now i got to go finish that pesky area of a triangle one thanks a ton for helping me lol, id be up much later if i was trying to go solo :/ back on the topic of the first question, u set 1/2(1-3x^2) = 0 so i should end up with 1/2 - 3/2x^2 = 0 and all i need to do is solve...

lol sorry, i was just doing a step twice, once on paper once in my head :P so min(2,-4) max(-2,4) correct? EDIT: lol u beat me to it

so what do I do with +/- 2, are these the y coords for my min/max? so all i would have to do is find the x value?EDIT: errr sorry, x coords for the min/max so id have to find the y values? which would just be: max: (2,2) min(-2,-2) ?

neither 2 or -2 make it 0 though.

wait wait wait duh, can't i just go x^2=4 and my roots are +/- 2 ??