Extreme and Intermediate value theorem

  • Thread starter Thread starter mikael27
  • Start date Start date
  • Tags Tags
    Theorem Value
Click For Summary
SUMMARY

The discussion focuses on proving that if the set S = {f(x) | a ≤ x ≤ b} contains more than one element for a continuous function f: [a, b] → R, then S is an interval of the form [c, d]. The Extreme Value Theorem (EVT) is applied to establish that the function attains its minimum and maximum values at points c1 and c2, respectively. Subsequently, the Intermediate Value Theorem (IVT) is utilized to demonstrate that all values between f(c1) and f(c2) are also included in S, confirming that S is indeed an interval.

PREREQUISITES
  • Understanding of the Extreme Value Theorem (EVT)
  • Familiarity with the Intermediate Value Theorem (IVT)
  • Knowledge of continuous functions on closed intervals
  • Basic concepts of real analysis
NEXT STEPS
  • Study the proofs of the Extreme Value Theorem (EVT)
  • Explore applications of the Intermediate Value Theorem (IVT)
  • Investigate properties of continuous functions on closed intervals
  • Learn about the implications of continuous mappings in real analysis
USEFUL FOR

Mathematics students, educators, and anyone studying real analysis or calculus, particularly those interested in the properties of continuous functions and their applications in mathematical proofs.

mikael27
Messages
59
Reaction score
0

Homework Statement



Let f : [a; b] ! R be an arbitrary continuous function. Let S = {f(x)| a<= x<=b}. Show
that if S contains more than one element, then S is an interval of the form [c, d].

Hint: First apply the Extreme Value theorem, then the Intermediate Value theorem.

Homework Equations





The Attempt at a Solution



dont have any clue

 
Physics news on Phys.org
If we suppose S contains more than one points then are a \le c_1 &lt; c_2 \le b such that f(c_1) \neq f(c_2). Now the EVT can be applied to say something about the relationship of these. Once that's established the MVT will show that it must be an interval.

This actually says something quite important about continuous mappings over real numbers.
 
In response to your PM, the EVT can be applied to say that, in addition (without loss of generaltiy) c_1 and c_2 are the minimum and maximum on this interval, respectively. We can say this because continuous function must attain their maximum and minimum.

Now the IVT can be applied to show that f also attains all values between f(c_1) and f(c_2). You can do the same for the intervals [a,c_1] and [c_2, b], then you have f(c_2) = d &gt; c = f(c_1) and so f([a,b]) = [c,d].


It stil needs some details, but that's the gist of it.
 

Similar threads

Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K