Write an equation of each horizontal tangent line to the curve.

[lude1]

Homework Statement

Write an equation of each horizontal tangent line to the curve.

Homework Equations

y = 2y^3 + 6x^2y - 12x + 6y = 1
y' = (4x - 2xy) / x^2 + y^2 + 1)

The Attempt at a Solution

Well, horizontal tangent line means the derivative equals zero. Thus,

4x - 2xy = 0
2x(2-y) = 0
x = 0, y = 2​

Since I need an equation, I plug x = 0 back into the original function and end up with

2y^3 + 6y = 1​

Then I need to solve for y. I guess this is more of an algebra question than a calculus question? But nevertheless, I don't know how to solve for y.. I could do 2y(y^2 + 3) = 1 but that doesn't help me.

 

[tiny-tim]

Hi lude1!

(try using the X² tag just above the Reply box )

2y^3 + 6x^2y - 12x + 6y = 1
y' = (4x - 2xy) / x^2 + y^2 + 1)

Where does the 4x come from?

(Or should the 12x be 12x² ?)

 

[lude1]

Ah, yes, you are correct! It should be 12x²

Thus:

y = 2y³ + 6x²y - 12x² + 6y = 1

and

y' = (4x - 2xy) / (x² + y² + 1)

 

[tiny-tim]

Ah, yes, you are correct! It should be 12x²

Thus:

y = 2y³ + 6x²y - 12x² + 6y = 1

(why do you keep starting with "y =" ? )

hmm … then, sorry, I don't think your 2y³ + 6y = 1 has a straightforward answer

(unless the = 1 should be = 8)

oh, and don't forget you still have to deal with the other case, y = 2.