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Calculus A help please
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dy/dx= (3xy - y^2) / (2xy - x^3) For vertical slope 1/0 dy/dx= (3xy - y^2) / (2xy - x^3) =1/0 (2xy - x^3)=0 x(2y-x^2)=0 x=0 y=x^2/2 x(x^2/2)^2 -x^3(x^2/2) = 6 x^5/4-x^5/2 =6 -x^5/4=6 x^5=-24 x=-0.935 y=0.437 Source: http://www.mathskey.com/question2answer/
Substitute y = x^2/2
(0,0) and (-0.935,0.437 ) are answers for c only
for horizontal dy/dx=0
for horizontal slope is zero so dy/dx =0
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dy/dx = (3xy - y^2) / (2xy - x^3) =0 3xy - y^2=0 y(3x-y)=0 y =0 y =3x xy^2 -x^3y = 6 x(3x)^2 -x^3(3x) = 6 3x^3-3x^4-6=0 solve this equation
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