Implicit function solution for $ y=f(x)$

Question

What does it mean with find the implicit solution to $y=f(x)?$ I know I'm supposed to check conditions and then use IVT to say $y$ can be solved uniquely for $x$. But what do they mean with solution, do I need to get the partial derivative of $G(x) = F(x,f(x))$ with respect to $x$ and how does it constitute a solution?

exercise

Answer 1

You don't need a theorem to find the solutions of this quadratic equation in $y$:

$$y=\pm x\sqrt{x-1}.$$

Then it is an easy matter to check that a) is the $+$ branch and b) the $-$ one.

Answer 2

You have to find a solution of the implicit equation $$ x^2+y^2-x^3 =0 $$ In general, if you have an implicit equation like $F(x,y)=0$, finding a solution means that you have to find that function $f$ such that $F(x,f(x))=0$. Now think about the circle, $F(x,y) = x^2+y^2 =0$. It is impossible to find a single $f$ such that $y=f(x)$ draws a full circle, but you can find such a function $f_1$ for the lower part of the circle ("near the south pole") and an $f_2$ for the upper part of the circle (close to the "north pole").

Regarding your specific problem, $$ y=f_1(x) = (-x^2+x^3)^{1/2} $$ or $$ y=f_2(x) = -(-x^2+x^3)^{1/2} $$ ...you just have to choose which one is appropriate close to the points $A$ and $B$ of your exercise.