Editing Implicit function theorem (section)

== First example ==
[[Image:Implicit circle.svg|thumb|right|200px|The unit circle of implicit equation {{math|1= ''x''<sup>2</sup> + ''y''<sup>2</sup> – 1 = 0}} cannot be represented as the graph of a function. Around the point {{math|'''A'''}} where the tangent is not vertical, the bolded [[circular arc]] is the graph of some function of {{mvar|x}}, while around {{math|'''B'''}}, there is no function of {{mvar|x}} with the circle as its graph.<br>This is exactly what the implicit function theorem asserts in this case.]]
If we define the function {{math|1=''f''(''x'', ''y'') = ''x''<sup>2</sup> + ''y''<sup>2</sup>}}, then the equation {{math|1=''f''(''x'', ''y'') = 1}} cuts out the [[unit circle]] as the [[level set]] {{math|1={(''x'', ''y'') {{!}} ''f''(''x'', ''y'') = 1}<nowiki/>}}. There is no way to represent the unit circle as the graph of a function of one variable {{math|1=''y'' = ''g''(''x'')}} because for each choice of {{math|''x'' ∈ (−1, 1)}}, there are two choices of ''y'', namely <math>\pm\sqrt{1-x^2}</math>.

However, it is possible to represent ''part'' of the circle as the graph of a function of one variable. If we let <math>g_1(x) = \sqrt{1-x^2}</math> for {{math|−1 ≤ ''x'' ≤ 1}}, then the graph of {{math|1=''y'' = ''g''<sub>1</sub>(''x'')}} provides the upper half of the circle. Similarly, if <math>g_2(x) = -\sqrt{1-x^2}</math>, then the graph of {{math|1=''y'' = ''g''<sub>2</sub>(''x'')}} gives the lower half of the circle.

The purpose of the implicit function theorem is to tell us that functions like {{math|''g''<sub>1</sub>(''x'')}} and {{math|''g''<sub>2</sub>(''x'')}} [[List of mathematical jargon#almost all|almost always]] exist, even in situations where we cannot write down explicit formulas. It guarantees that {{math|''g''<sub>1</sub>(''x'')}} and {{math|''g''<sub>2</sub>(''x'')}} are differentiable, and it even works in situations where we do not have a formula for {{math|''f''(''x'', ''y'')}}.