I need to show that: $ \forall x\in \Re \quad \exists !\quad y=y(x)$ in order that $ \ { y }^{ 3 }+y-{ x }^{ 3 }+x=0$ and than i want to determine all local Extremas of $ y=y(x)$.
It does look to me that the first Part is a Problem of differential eqautions however i haven't studied those yet. We can see that the function is somehow symetric which would answer the Problem but I dont think thats it. So what do I need to do to identify $ y=y(x)$ and prrof the first part?
Edit: using the implicit function theorem: $f(x,y)={ y }^{ 3 }+y-{ x }^{ 3 }+x\\ f(x,y(x))={ y }^{ 3 }+y-{ x }^{ 3 }+x=0\\ \\ { y(x) }^{ 3 }+y(x)-{ x }^{ 3 }+x=0\\ { 2y(x) }^{ 3 }={ x }^{ 3 }-x\\ y(x)^{ 3 }=\frac { { x }^{ 3 }-x }{ 2 } \\ y(x)=\sqrt [ 3 ]{ \frac { { x }^{ 3 }-x }{ 2 } } $ would this approach be right? Dont know if i used the theorem correctly first time i used it
Let $f(x,y)=y^3 +y -x^3 +x$. Then $f_y'\ge 1\ne 0$. By the implicit function theorem, for any $(x_0,y_0)$ such that $f(x_0,y_0)=0$, there is a local implicit function $y=h(x)$ from $x_0$ to $y_0$ such that $f(x,h(x))=0$, and we have $$h'(x_0)=-f_x'(x_0,y_0)/f_y'(x_0,y_0)=-\frac{-3x_0^2+1}{3h(x_0)^2+1}.$$ By the arbitrary choice of $x_0$, the above expression is indeed global. This indicates that the implicit function $h$ attains its local extreme values at $x=\pm\sqrt{\dfrac13}$.