Editing Rolle's theorem (section)

== Examples ==

===Half circle===
[[File:semicircle.svg|thumb|300px|A semicircle of radius {{mvar|r}}]]
For a radius {{math|''r'' > 0}}, consider the function
<math display="block">f(x)=\sqrt{r^2 - x^2},\quad x \in [-r, r].</math>

Its [[graph of a function|graph]] is the upper [[semicircle]] centered at the origin. This function is continuous on the closed interval {{closed-closed|−''r'', ''r''}} and differentiable in the open interval {{open-open|−''r'', ''r''}}, but not differentiable at the endpoints {{math|−''r''}} and {{mvar|r}}. Since {{math|1=''f&thinsp;''(−''r'') = ''f&thinsp;''(''r'')}}, Rolle's theorem applies, and indeed, there is a point where the derivative of {{mvar|f}} is zero. The theorem applies even when the function cannot be differentiated at the endpoints because it only requires the function to be differentiable in the open interval.

===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).

===Functions with zero derivative===

Rolle's theorem implies that a [[differentiable function]] whose derivative is {{tmath|0}} in an interval is constant in this interval. 

Indeed, if {{mvar|a}} and {{mvar|b}} are two points in an interval where a function {{mvar|f}} is differentiable, then the function
<math display=block>g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)</math>
satisfies the hypotheses of Rolle's theorem on the interval {{tmath|[a,b]}}.

If the derivative of {{tmath|f}} is zero everywhere, the derivative of {{tmath|g}} is
<math display=block>g'(x)=\frac{f(b)-f(a)}{b-a},</math>
and Rolle's theorem implies that there is {{tmath|c\in (a,b)}} such that
<math display=block>0=g'(c)=\frac{f(b)-f(a)}{b-a}.</math>

Hence, {{tmath|1=f(a)=f(b)}} for every {{tmath|a}} and {{tmath|b}}, and the function {{tmath|f}} is constant.