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 )
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...
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) ?