What is local minimum and maximum of implicit function $F(x,y)=x^4+y^4−x^2 −y^2=0$?
I calculated first and second derivative:
$$\frac{\partial F(x, y(x))}{\partial x}=4x^3+4y^3y'-2x-2yy'$$
$$\frac{\partial F^2(x, y(x))}{\partial x^2}=12x^2+12y^2(y')^2+12y^3y''-2(y')^2-2yy''-2$$
$$y'=\frac{2x-4x^3}{4y^3-2y}$$
$$y''=\frac{(y')^2-6x^2-6y^2(y')^2+1}{6y^3-y}$$
Then I don't know how to continue. How to find stationary points? How to decide if stationary point is local minimum or maximum?
Best Answer
When differentiating implicitly you should get that: \begin{align*} y'(x) = - \frac{\frac{\partial F}{\partial x}(x,y(x))}{\frac{\partial F}{\partial y}(x,y(x))} = \frac{2x-4x^3}{2y-4y^3} \end{align*} Now at an extremum we have that $y'(x)=0$. Lets assume that $2y-4y^3 \neq 0 $ (if we want $y$ to be an implicit function of $x$ then this is a demand). Then $y'(x)=0 \Leftrightarrow 2x-4x^3 =0 \Leftrightarrow 2x(1-2x^2)=0 \Leftrightarrow x =0 \vee x= \pm \sqrt{\frac{1}{2}}$.
Now just check the values of $y$ at these $x$-values (and possible endpoints of the domain of definition of $y$).