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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 $y$! –  Mercy Sep 27 '12 at 20:55
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1 Answer 1

up vote 1 down vote accepted

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). –  Farmor Sep 27 '12 at 20:32
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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. –  rschwieb Sep 27 '12 at 20:40
    
You got it right... You have no idea what it is, and actually there is no restriction there... It could be anything. –  N. S. Sep 27 '12 at 20:41
    
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. –  rschwieb Sep 27 '12 at 20:41
    
Thanks very much. Yes, that was exactly the original function. –  Farmor Sep 27 '12 at 20:46
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