Editing Inverse function theorem (section)

== Counter-example ==
[[File:Inv-Fun-Thm-3.png|thumb|The function <math>f(x)=x+2 x^2\sin(\tfrac1x)</math> is bounded inside a quadratic envelope near the line <math>y=x</math>, so <math>f'(0)=1</math>. Nevertheless, it has local max/min points accumulating at <math>x=0</math>, so it is not one-to-one on any surrounding interval.]]
If one drops the assumption that the derivative is continuous, the function no longer need be invertible. For example <math>f(x) = x + 2x^2\sin(\tfrac1x)</math> and <math>f(0)= 0</math> has discontinuous derivative
<math>f'\!(x) = 1 -2\cos(\tfrac1x) + 4x\sin(\tfrac1x)</math> and <math>f'\!(0) = 1</math>, which vanishes arbitrarily close to <math>x=0</math>. These critical points are local max/min points of <math>f</math>, so <math>f</math> is not one-to-one (and not invertible) on any interval containing <math>x=0</math>. Intuitively, the slope <math>f'\!(0)=1</math> does not propagate to nearby points, where the slopes are governed by a weak but rapid oscillation.