Finding Points Where dy/dx of (y-2)^3 + y(x-6) = 0 Does Not Exist

[jay_jay_lp]

At what points on the curve (y-2)^3 + y(x-6) = 0 does dy/dx not exist?

So I've tried a few approaches:
1. find point where dx=0, since -Fx/Fy = dy/dx, therefore also Fy=0. So I took the partial derivative wrt y and got another equation, 3(y-2)^2 + (x-6) = 0. From here I don't know where to go.

2. Graphically:

[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP296019ha1c8h91aed5c300005983bc124ea5gai3?MSPStoreType=image/gif&s=10&w=200&h=205&cdf=Coordinates&cdf=Tooltips

From this it would seem that approximately when x<-20, there is no derivative or no defined derivative since there is more than one?

But how do I find this algebraically?

Thanks

Jay

 

[LCKurtz]

It is true that F(x,y) = (y-2)³ + y(x-6) = 0 does not define a single implicit function of x. There are three separate functions defined as your graph shows. What you are looking for is vertical tangent lines. As you have noticed, one such point is (6,2). What isn't so obvious is there is another point on the curve that gives a zero denominator. One way to find it is to solve F(x,y) = 0 together with your denominator F_y(x,y)=0. That will get points on the curve where the derivative doesn't exist.

Another way, which works in this particular equation, is to solve for x in terms of y in the original function, and calculate where dx/dy = 0.

 

[jay_jay_lp]

Thanks a lot,

I'll reply again if I need more help

Jay