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 ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
Let 
be continous on ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
and differentiable on 
Let 
Then 
such that 
Proof
The result is trivial for the case ![$f([a,b])=\{f(a)\}$](http://latex.artofproblemsolving.com/c/d/c/cdc660c1a05ec6f54a3a2f57a03a47e3eaa23649.png)
. Hence, let us assume that is a non-constant function.
Let ![$M=\sup\{f([a,b])\}$](http://latex.artofproblemsolving.com/3/b/b/3bbce4c453e5dbd5353e06c8c6e58b2e0d592bf3.png)
and ![$m=\inf\{f([a,b])\}$](http://latex.artofproblemsolving.com/1/b/c/1bcecd82871ae4cb9ccf9b2da9d2ec1bbf513980.png)
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
