Module 4 : Local / Global Maximum / Minimum and Curve Sketching
Lecture 11 : Absolute Maximum / Minimum [Section 11.1]
11.1.3
Theorem:
Let be continuous. Then, assumes its absolute maximum as well as its absolute minimum at some points which are either critical point of or are the end points .
Proof:
Since is continuous, there exists such that is absolute maximum and is absolute minimum. Suppose that neither of them is any one of the end points . Then, . Note that, is a point of local maximum and is a point of local minimum also. Now either is not differentiable at or if is differentiable at , then by lemma 9.1.4. Similarly, either is not differentiable at or if is differentiable at , then, , again by lemma 9.1.4.
be continuous. Then, 
assumes its absolute maximum as well as its absolute minimum at some points which are either critical point of 
or are the end points 
. 
is continuous, there exists 
such that 
is absolute maximum and 
is absolute minimum. Suppose that neither of them is any one of the end points 
. Then, 
. Note that, 
is a point of local maximum and 
is a point of local minimum also. Now either 
is not differentiable at 
or if 
is differentiable at 
, then 
by lemma 9.1.4. Similarly, either 
is not differentiable at 
or if 
is differentiable at 
, then, 
, again by lemma 9.1.4. 
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