Min-MaxTheorem:If a function is continuousin a closedinterval then f(x) has both a maximum
valueand a minimumvaluein In orderto understandthe prooffor the Min-MaxTheoremconceptually,
attemptto drawa functionon a closedinterval(includingthe endpoints)so that no pointis at the highest
part of the graph.No matterhow the functionis sketched,therewill be at leastone pointthat is highest.
We can now relateextremevaluesto derivativesin the followingTheoremby the Frenchmathematician
Fermat.
Theorem:If is an extremevalueof for someopenintervalof and if exists,then
Proof: The theoremstatesthat if we havea local max or local min, and if exists,then we musthave
Suppose that has a local max at Then we have for someopen interval
with
So
Consider.
Since , we have
Since exists,we have
and so.
If we take the left-handlimit, we get
Hence and it mustbe that
If is a local minimum,the sameargumentfollows.
Definition: We will call acriticalvaluein if or doesnot exist,or if
is an endpointof the interval.
We can now statethe ExtremeValue Theorem.
ExtremeValue Theorem: If a function is continuousin a closedinterval , with the maximum
of at and the minimumof at then and are criticalvaluesof
Proof: The prooffollowsfrom Fermat’s theoremand is left as an exercisefor the student.
Example1: