# Find the solution of y = f(x)

Find the solution $$y = f(x)$$ of: $$x^2 + y^2 - x^3 = 0$$ Near the following points $$(5,10)$$,$$(10,-30)$$

I think I need to use the implicit function theorem and I tried this

• First for the definition the function has continuous partial derivatives

• Applying $$(5,10)$$: hence, $$F(5,10) = 0$$

• $$\frac{\partial}{\partial y}(5,10) = 2\cdot 10 = 20 \neq 0$$

Hence by the implicit function theorem it has a unique solution in the form of $$y(x)$$ around $$(5,10)$$

But what is this solution?

• Oh yes I forgot to add = 0 Thankss @lulu May 21 '19 at 20:03
• What do you mean by solution? Do you want to express $y$ in terms of $x,$ or what? May 21 '19 at 20:18

The two solutions would be $$y = \sqrt{x^3 - x^2}$$ around $$(5,10)$$, and $$y = -\sqrt{x^3 - x^2}$$ around $$(10, -30)$$.
If $$x^2+ y^2- x^3= 0$$ then $$y^2= x^3- x^2$$ so $$y= \pm\sqrt{x^3- x^2}$$. Since 10 is positive the branch that passes through (5, 10) is $$y= \sqrt{x^3- x^2}$$. Since -30 is negative the branch that passes through (10, -30) is $$y= -\sqrt{x^3- x^2}$$.
Note that $$x^2 + y^2 - x^3 = 0 \implies y^2=x^3-x^2$$
Therefore you have two functions $$y=\pm \sqrt {x^3-x^2}$$
Therefore near the point $$(5,10)$$ you pick the positive sign and near the point $$(10,-30)$$ you pick negative sign.