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I have a function for which I have calculated:

$\dfrac{d}{dx}f(x,y)=0 $

and

$\dfrac{d}{dy}f(x,y)=2y+\cos(y)$

How can I proceed to calculate the critical points?

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    Looks like your function only depends on $y$!2012-09-27

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Just as with equations in one variable, determine when the partial derivatives become 0. Here, one is already zero so no information there...

But the other one is $2y+\cos(y)=0$. It looks like there isn't a closed form for the solution, so you'd need an approximation...

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    If I solve this with the calculator I get -0.45. What will the critical points be? (?,-0.45) I don't understand how to get the x value(s).2012-09-27
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    Since there are no restrictions on $x$, $(x,-0.45)$ is a critical point *for all* $x$! That is, The entire horizontal line consists of critical points.2012-09-27
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    You got it right... You have no idea what it is, and actually there is no restriction there... It could be anything.2012-09-27
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    Was your original $f(x,y)=y^2+\sin(y)$? If it was, you can see that in 3-d, it's the same shape at every $x$ cross-section.2012-09-27
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    Thanks very much. Yes, that was exactly the original function.2012-09-27
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    An $x$ cross-section of this looks like a warped parabola... the critical points you are picking up sit at the quasi-vertex of these quasi-parabolas.2012-09-27
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    From the 3D-Graph I can see that the critical points are minimum. But with the formula $f_{12}(x,y))^2-f_{11}(x,y))*f_{22}(x,y))=0$ I get it is undefined. I suppose the correct answer to the question classification will be they are minimums.2012-09-27
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    You should use partial derivative notation $\frac{\partial}{\partial x} f(x,y) = 0$, etc.2012-09-27