Posted by Anonymous on Wednesday, March 13, 2013 at 9:12pm.
a)
If x3 + y3  xy2 = 5, find dy/dx.
(b)
Using your answer to part (a), make a table of approximate yvalues of points on the curve near x = 1, y = 2. Include x = 0.96, 0.98, 1, 1.02, 1.04.
(c)
Find the yvalue for x = 0.96 by substituting x = 0.96 in the original equation and solving for y using a computer or calculator. Compare with your answer in part (b).
(d)
Find all points where the tangent line is horizontal or vertical.

calculus  Reiny, Wednesday, March 13, 2013 at 9:26pm
a) Assuming you meant
x^3 + y^3  x y^2 = 5
3x^2 + 3y^2 dy/dx  x(2y)dy/dx  y^2 = 0
dy/dx(3y^2  2xy) = y^2  3x^2
dy/dx = (y^2  3x^2)/(3y^2  2xy)
b) just arithmetic, you do it
c) yours
d) for a horizontal tangent, dy/dx = 0
then y^2  3x^2 = 0
y^2 = 3x^2
y = ± x√3
sub into the first:
x^3 + 3√3x^3  x(x√3) = 5
x^3 + 3√3 x^3  √3 x^2  5 = 0
Nastiness!!!!
wolfram says x = 1.03412 correct to 5 decimals
now sub that into the original and find y, equally nasty.
for vertical asymptote , denominator of dy/dx = 0
3y^2  2xy) = 0
y(3y  2x) = 0
y = 0 , that is not bad
or
y = 2x/3
more mess ahead
I suggest you use the Wolfram page or your builtin equation solver of your fancy calculator to solve these equations.
(it might be a good idea to check my work, lately I have been making silly errors by not writing out the solution first)
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