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algebra & calculas problem?
f(x,y) = 6x^2 - 2x^3 + 3y^2 + 6xy
fx = 12x - 6x^2 + 6y
fy = 6y + 6x
fy = 0 ; y = -x
fx = 0 ; 12x - 6x^2 + 6y = 0
now i'm wondering should i divide 6 or - 6 into fx because when i substitute y into x ,it would give the same critical points but the
graph aren't the same x(1-x) & x(x-1) or it does not matter?
one more thing , i'm not so sure on how to tell the the points in
the equation without using a calculator , i mean like x(1-x)
x = 0 , x =1 ?
1 Answer
- icemanLv 71 decade agoFavorite Answer
12x - 6x^2 + 6y = 0 >>> divide by - 6
x^2 - 2x = y >>> fy = 0 y = - x
x^2 - 2x = - x >>> x^2 - x = 0
x(x - 1) = 0 >>> Answer to your 2ND question,
if ab = 0 then both a and be could be = 0 so a = 0 or b = 0
x(x - 1) = 0
x = 0 ,
x - 1 = 0 >>> x = 1
