Discussion
Consider the function \(f : \mathbf{R}^{2} \to \mathbf{R}\) given by \(f(x, y) = x^{2} + y^{2} - 4y\). Then in a neighbourhood of (0, 0) :
Answer: B
Near (0, 0), \(y\) can be expressed as a function of \(x\).
Consider the level curve \(f(x, y) = f(0, 0) = 0\). At the origin, \(f_{x} = 2x = 0\) and \(f_{y} = 2y - 4 = -4 \neq 0\). By the implicit function theorem, the non-zero \(f_{y}\) lets us solve for \(y\) as a smooth function of \(x\) near 0. Indeed \(y = 2 - \sqrt{4 - x^{2}}\) near the origin.
\(x\) cannot be written as a function of \(y\): solving gives \(x = \pm\sqrt{4y - y^{2}}\), two values for each small \(y \gt 0\), and none for \(y \lt 0\). So A and D fail, and C fails because \(y\) does work.
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