Editing Rolle's theorem (section)

===Absolute value===
[[File:Absolute value.svg|thumb|300px|The graph of the absolute value function]]
If differentiability fails at an interior point of the interval, the conclusion of Rolle's theorem may not hold. Consider the [[absolute value]] function
<math display="block">f(x) = |x|,\quad x \in [-1, 1].</math>

Then {{math|1=''f&thinsp;''(−1) = ''f&thinsp;''(1)}}, but there is no {{mvar|c}} between −1 and 1 for which the {{math|''f&thinsp;''′(''c'')}} is zero. This is because that function, although continuous, is not differentiable at {{math|1=''x'' = 0}}. The derivative of {{mvar|f}} changes its sign at {{math|1=''x'' = 0}}, but without attaining the value 0. The theorem cannot be applied to this function because it does not satisfy the condition that the function must be differentiable for every {{mvar|x}} in the open interval. However, when the differentiability requirement is dropped from Rolle's theorem, {{mvar|f}} will still have a [[critical number]] in the open interval {{open-open|''a'', ''b''}}, but it may not yield a horizontal tangent (as in the case of the absolute value represented in the graph).