Editing Rolle's theorem (section)

== Proof of the generalized version ==

Since the proof for the standard version of Rolle's theorem and the generalization are very similar, we prove the generalization.

The idea of the proof is to argue that if {{math|1=''f&thinsp;''(''a'') = ''f&thinsp;''(''b'')}}, then {{mvar|f}} must attain either [[maxima and minima|a maximum or a minimum]] somewhere between {{mvar|a}} and {{mvar|b}}, say at {{mvar|c}}, and the function must change from increasing to decreasing (or the other way around) at {{mvar|c}}. In particular, if the derivative exists, it must be zero at {{mvar|c}}.

By assumption, {{mvar|f}} is continuous on {{closed-closed|''a'', ''b''}}, and by the [[extreme value theorem]] attains both its maximum and its minimum in {{closed-closed|''a'', ''b''}}. If these are both attained at the endpoints of {{closed-closed|''a'', ''b''}}, then {{mvar|f}} is [[constant function|constant]] on {{closed-closed|''a'', ''b''}} and so the derivative of {{mvar|f}} is zero at every point in {{open-open|''a'', ''b''}}.

Suppose then that the maximum is obtained at an [[interior point]] {{mvar|c}} of {{open-open|''a'', ''b''}} (the argument for the minimum is very similar, just consider {{math|−''f&thinsp;''}}). We shall examine the above right- and left-hand limits separately.

For a real {{mvar|h}} such that {{math|''c'' + ''h''}} is in {{closed-closed|''a'', ''b''}}, the value {{math|''f&thinsp;''(''c'' + ''h'')}} is smaller or equal to {{math|''f&thinsp;''(''c'')}} because {{mvar|f}} attains its maximum at {{mvar|c}}. Therefore, for every {{math|''h'' > 0}},
<math display="block">\frac{f(c+h)-f(c)}{h}\le0,</math>
hence
<math display="block">f'(c^+):=\lim_{h \to 0^+}\frac{f(c+h)-f(c)}{h}\le0,</math>
where the limit exists by assumption, it may be minus infinity.

Similarly, for every {{math|''h'' < 0}}, the inequality turns around because the denominator is now negative and we get
<math display="block">\frac{f(c+h)-f(c)}{h}\ge0,</math>
hence
<math display="block">f'(c^-):=\lim_{h \to 0^-}\frac{f(c+h)-f(c)}{h}\ge0,</math>
where the limit might be plus infinity.

Finally, when the above right- and left-hand limits agree (in particular when {{mvar|f}} is differentiable), then the derivative of {{mvar|f}} at {{mvar|c}} must be zero.

(Alternatively, we can apply [[Fermat's theorem (stationary points)|Fermat's stationary point theorem]] directly.)