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If $\frac{\partial f}{\partial x}= 0$ when $x = 1$, and $\frac{\partial f}{\partial x} = (2y-x)$

Do we have to replace $x$ by 1? to get the critical coordinate $y$?

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Indeed you have :) – Pragabhava Dec 6 '12 at 0:20

The critical points occur when $f_{x}=f_{y}=0$. So, necessarily, any critical point must occur when $x=1$ so that we obtain $2y-1=0$ and $y=\frac{1}{2}$ as desired. So you are correct.

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Note that just because $(1, \frac{1}{2})$ is a potential critical point doesn't mean it is one--you still need to check $f_y$ – apnorton Dec 6 '12 at 0:34
@anorton No, the definition of a critical point is that the partial derivatives are zero. However, just because it is a critical point does not mean that it is a maximum or minimum, which might be what you are referring to. – Jebruho Dec 6 '12 at 0:55
I disagree... My textbook ("University Calculus" by Hass, Weir, and Thomas) defines a critical point as "An interior point of the domain of a function $f(x,y)$ where both $f_x$ and $f_y$ are zero or where one or both of $f_x$ and $f_y$ do not exist is a critical point of $f$." Wikipedia also verifies this:… – apnorton Dec 6 '12 at 1:26
@anorton That's what I'm saying. That's not what you said. I don't know what you mean by a "potential critical point." – Jebruho Dec 6 '12 at 1:27
Oh. Sorry. I meant it has the potential to be a critical point. It could be, but we don't know yet. To verify, we must know $f_y$. – apnorton Dec 6 '12 at 1:29

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