Question on extremums, can anyone provide some help?

Question

The real question: Let $f \colon \mathbb{R}^3 \to \mathbb{R}$; $f(x,y_1,y_2)=x^3−xy_1y_2+y_2^2−16$. Show that there exists a differentiable real function $g$ so that in some neighboorhood $(1,4)$ $$f(g(y_1,y_2),y_1,y_2)=0 $$Find $g′(1,4)$ and ${\partial^2g\over \partial x^2}(1,4).$ My theorem of implicit function is:$$$$ $$\large {Implicit\ Theorem:}$$ $$\begin {cases} a.)X,Y,Z ;Banach- spaces, W\subset X \times Y. \\ b.)F:W \to Z; F\in C^1; F(a,b)=0.\\ c.)D_yF(a,b):Y\to Z -isomorphism-Y-and-Z \end {cases}$$ $Conclusions:$ $$There\ exists\ an \ open\ neighboorhood\ U(a)\in X,\ an \ open \ neighboor\ W'=W'(a,b)\subset W\subset X \times Y and\ a\ function\ f:U\to Y, that: $$ $$\begin {cases} 1.)(x,y)\in W';F(x,y)=0 \iff x\in U ;y=f(x);f(a)=b.\\ 2.)f\in C^1;f'(a)=-D_yF^{-1}\circ D_xF. \\ 3.)If\ F \in C^k \implies f\in C^k,(k\geq1). \end {cases}$$

Answer 1

There are in fact three different such functions $g$.

Preliminary remark: The letters $X$ and $Y$ in your formulation of the implicit function theorem do not correspond to the names $x$, $y_1$, $y_2$ of the variables in your example!

We are given the function $$f(x,y,z):=x^3-xyz+z^2-16$$ and are interested in the "surface" $S: \>f(x,y,z)=0$ in ${\Bbb R}^3$. To be exact, we are also given the point $P:=(1,4)$ in the $(y,z)$-plane, and are asked to present a suitable part of $S$ as a graph $$x=g(y,z)\qquad\bigl((y,z)\in U\bigr)\tag{1}$$ with $g$ defined in some neighborhood $U$ of $P$. In order to find points of $S$ projecting to $P$ we have to solve the equation $$\phi(x):=f(x,1,4)=x^3-4x=0$$ for $x$. It follows that $x\in\{-2,0,2\}$, which means that three different points of $S$ project to $P$. In the following we consider the point $(2,1,4)\in S$; the other two can be handled analogously.

Since $f_x(2,1,4)=\phi'(2)=4\ne0$ the implicit function theorem guarantees the following: There is an open box $W:=V\times U\subset{\Bbb R}\times{\Bbb R}^2$ with center $(2,1,4)$ (a "window") such that $S\cap W$ has a representation of the form $(1)$ with a $g\in C^1(U,V)$ and $g(1,4)=2$. This implies $$f\bigl(g(y,z),y,z\bigr)\equiv0\qquad\bigl((y,z)\in U\bigr)\ .\tag{2}$$ We now take in $(2)$ the partial derivative with respect to $y$. Using the chain rule we obtain $$f_x\bigl(g(y,z),y,z\bigr)g_y(y,z)+f_y\bigl(g(y,z),y,z\bigr)\cdot 1\equiv0\ ,$$ and putting $(y,z):=(1,4)$ we obtain $$f_x(2,1,4)g_y(1,4)+f_y(2,1,4)=0\ ,$$ so that $$g_y(1,4)=-{f_y(2,1,4)\over f_x(2,1,4)}=-{-8\over 4}=2\ .$$ In a similar way, taking the partial derivative with respect to $z$, we obtain from $(2)$ the identity $$f_x\bigl(g(y,z),y,z\bigr)g_z(y,z)+f_z\bigl(g(y,z),y,z\bigr)\cdot 1\equiv0\ ,$$ which for $(y,z):=(1,4)$ gives $$g_z(1,4)=-{f_z(2,1,4)\over f_x(2,1,4)}=-{6\over4}=-{3\over2}\ .$$ What you call $g'(1,4)$ therefore can be written as $$dg(1,4).(Y,Z)=2Y-{3\over2}Z\ .$$ The expression ${\partial^2 g\over \partial x^2}(1,4)$ in your question does not make sense..

Answer 2

I'll change the notations for better comprehension. $$F(y_1,y_2,x)=x^3−xy_1y_2+y_2^2−16$$ $a=(1,4)=(y_1,y_2)\in X=R^2 ; b=0=x\in Y=R;Z=R$ (Z and Y must be the same dimensions..because of isomorphism condition in theorem)Condition $1.\ and\ 2.$ is satisfied obviously because real spaces are Banach spaces and the partial derivatives of $F$ also $\in C^1$ and $F(a,b)=0$. So that means (by the theorem) there exists $g(a)=b; g\in C^1$ such that :$g'(a)=-(D_yF(a,b))^{-1}\circ D_xF(a,b)=...-{4}$