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here is the question:

Can the surface whose equation is $xy-y \log (z)+\sin (xz)=0$ be represented in form $z=f(x,y)$ near $(0,2,1)$

If $z$ is going to be one, wouldn't that automatically make it $f(x,y)$ ? Can it really be that simple?

thanks for hints/help guys.

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It says near $(0,2,1)$, so $z$ will be close to $1$ but not necessarily equal to $1$. Have you learned the implicit function theorem? –  Henry T. Horton Mar 28 '13 at 17:56
    
maybe a long time ago. thanks for the heads up I will look at it. –  Neo Mar 28 '13 at 18:02

1 Answer 1

up vote 0 down vote accepted

so I think I have a solution, could someone check it ?

$F(x,y,z) = xy-ylog(z)+sin(xz)$

$F_z = xcos(xz) - \frac{y}{z} =>(0,2,1) = -2 \not =0$ Therefore by IFT there is a form $z=f(x,y)$ near point $(0,2,1)$

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