find all points where tangent is horizontal

[math]

xy^3 - 3x^2 + y^3 = 0

a. Find second derivative at x=2
I got this. dy/dx = (-y^2 + 6x)/(2xy + 3y^2)
Since x=2, I plugged 2 into original function to solve for y, and i got y=1.781
I plugged in x=2 and y=1.781 into dy/dx and got dy/dx=.53
Then I used quotient rule to find 2nd derivative, then plugged in values for x, y, and dy/dx and got second deriv = - .115

Now this is where I am stuck, I included the above since its necessary for part b:
b. Find all points where tangent is horizontal.
I would think to set dy/dx equal to zero, but then I have -y^2 + 6x = 0 . I'm not sure where to go from here.

Thanks!

 

[tkhunny]

You may wish to take another shot at it.

Did you notice that \(\displaystyle \L\;y^{3}\;=\;\frac{3x^{2}}{x+1}\)?

For x = 2, this makes \(\displaystyle \L\;y\;=\;2^{2/3}\), not quite the 1.781 you managed.

This makes the first derivative substantially simpler.

\(\displaystyle \L\;y'\;=\;\frac{x}{y}*\frac{x+2}{(x+1)^{2}}\)

The second derivative is a bit more tedious, so of course I'll leave that for you. :evil:

Remember, the tangent is horizontal where the Numerator is zero AND the denominator is NOT zero AND everything has to exist AND the second derivative needs to behave, too. The first derivative is only a hint. I'm thinking x = 0 and x = -2 are pretty obvious choices, but I wonder what the associated y-values might be...