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?
2006-12-11
13:04:31
·
3 answers
·
asked by
Anonymous
in
Mathematics