Horizontal tangent lines?

Question

f(x) = 3x^(4)y - 7x^4 - 2y + 4y^3 = -3

-12x^(3)y + 28x^3
---------------------- = dy/dx
3x^4 - 2 + 12y^2

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

Find the equation of the normal line at the point x = 0.

------------------------------...

For the first part I set the top part of dy/dx to 0.
-12x^(3)y + 28x^3 = 0
and I got y = 7/3
But when I go back to f(x) and substitute 7/3 for y, the function wont work. Is the answer no horizontal tangents? I think I did something wrong.

For the second part I substituted 0 into f(x) and found that y = 1.09, so I get the point (0,-1.09)
I plug those into dy/dx and I get 0. So I get a tangent slope of 0.... this contridicts what I got in the first part.

What did I do wrong?

Answer 1

you got it wrong
dy/dx = 12x^(3)y - 28x^3

Answer 2

At y = 7/3, f(x) does not depend on x, and has a constant value of y which is the solution of the equation -2y+4y^3+3 = 0. So the equation of the line is slmply y = r where r is the solution (root) of that equation (as you found, r=1.09). dy/dx at x = 0 is 0, and the line normal to that is the vertical line x = 0; that is its equation.

Answer 3

f(x) = 3x^(4)y - 7x^4 - 2y + 4y^3 = -3
f'(x)= 3x^(4)(y') + y*12x^3) - 28x^3 -2y' + (12y^2 * y')
0 = 3x^(4)(y') + y*12x^3 - 28x^3 -2y' + (12y^2 * y')
2y' -3x^(4)(y') -(12y^2 * y')= y*12x^3 - 28x^3
y'( 2-3x^4 -12y^2) =y*12x^3 - 28x^3
y' = (12yx^3 -28x^3)/ (2- 3x^4 -12y^2)
that's what i'm getting for the first deriv? Eh. Calc I was 2 years ago.

okay so now setting the numerator equal to zero...
12yx^3 -28x^3 = 0
x^3(12y-28)=0
x=0 12y=28 (y=7/3).
eh, horizontal tangents have a slope of zero. yes. we've establishted that. so that means that in the form (y-y1)=m(x-x1) the right side of the equation is zero. yes. so y=said number.
so shouldn't said number be 7/3? i think so. so horiz tangents at y=7/3.