Implicit function theorem for $f(x,y,z)=z^2x+e^z+y$

Question

$f: mathbb{R}^3 rightarrow mathbb{R}$, $f(x,y,z)=z^2x+e^z+y$

Show that an neighboorhood $V$ of $(1,-1)$ in $mathbb{R}^2$ and a continuous differentiable function $g:V rightarrow mathbb{R}$ with $g(1,-1)=0$ and $f(x,y,g(x,y))=0$ for $(x,y) in V$ exists.

Calculate $D_1g(1,-1)$ and $D_2g(1,-1)$.

$nabla f(x,y,z)=begin{pmatrix}
f_x\
f_y\
f_z
end{pmatrix}=begin{pmatrix}
z^2\
1\
e^z+2zx
end{pmatrix}$
Can someone help me with this?

Fred · Accepted Answer

Show that $f(1,-1,0)=0$ and $f_z(1,-1,0) ne 0$.The implicit-function - theorem then shows that 1. holds.
For $(x,y) in V$ we then have
$$(*) quad 0=xg(x,y)^2+e^{g(x,y)}+y.$$
Differentiate $(*)$ with respect to $x$, plug in $(x,y)=(1,-1)$, then you can compute $D_1g(1,-1).$
Differentiate $(*)$ with respect to $y$, plug in $(x,y)=(1,-1)$, then you can compute $D_2g(1,-1).$

Implicit function theorem for $f(x,y,z)=z^2x+e^z+y$

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