Difference between revisions of "Rolle's Theorem"
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The result is trivial for the case <math>f([a,b])=\{f(a)\}</math>. Hence, let us assume that <math>f</math> is a non-constant function. | The result is trivial for the case <math>f([a,b])=\{f(a)\}</math>. Hence, let us assume that <math>f</math> is a non-constant function. | ||
− | Let <math>M=\sup\{f([a,b]\}</math> and <math>m=\inf\{f([a,b]\}</math> | + | Let <math>M=\sup\{f([a,b])\}</math> and <math>m=\inf\{f([a,b])\}</math> |
Without loss of generality, we can assume that <math>M\neq f(a)</math> | Without loss of generality, we can assume that <math>M\neq f(a)</math> | ||
Revision as of 21:54, 14 February 2008
Rolle's theorem is an important theorem among the class of results regarding the value of the derivative on an interval.
Statement
Let
Let be continous on and differentiable on
Let
Then such that
Proof
The result is trivial for the case . Hence, let us assume that is a non-constant function.
Let and Without loss of generality, we can assume that
By the Extremum value theorem, such that
Assume if possible
Let
Hence, such that
i.e. ,
Thus we have that if , contradicting the assumption that is a maximum.
Similarly we can show that leads to contradiction.
Therefore,
QED