y^2 = x^2 + 4/x
2y y' = 2x - 4/x^2
y' = (2x^3-4)/x^2 * 1/2y = (x^3-2) / x^2y
Now as long as y≠0, y'=0 when x=∛2
y(∛2) = ±√(∛4+4/∛2) = ±√(6/∛2)
so that's ok.
It looks like the only place with horizontal tangents is at (∛2,6/∛2), so the equation of the line is
y = 6/∛2
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y+%3D+x^2+%2B+4%2Fx%2C+y%3D6%2F%E2%88%9B2+for+-6+%3C%3D+x+%3C%3D+3
Find the equation of ALL horizontal tangents to the curve y^2=x^2+4/x, if any exists.
My work: Derivative- First I multiply across and got x^2+4=xy^2
2x+0=(x)(2y(dy/dx)+(y^2)
2x-y^2=2xy(dy/dx)
dy/dx=2x-y^2/2xy=x-y^2/xy
I don't know what to do next... Can you help me? Thanks :D
1 answer
1 answer