Intermediate Value Theorem question

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Homework Help Overview

The discussion revolves around the properties of a continuous function defined on a closed interval [a,b], specifically focusing on the conditions under which the function is strictly increasing. The original poster presents a scenario involving the Intermediate Value Theorem (IVT) and questions the implications of the function's continuity and distinct values at different points.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the IVT to demonstrate the existence of intermediate values within the interval but expresses confusion regarding the implications of their example, which they believe contradicts the theorem's assumptions. Other participants point out that the provided example violates the condition of distinct function values for different inputs.

Discussion Status

Participants are actively engaging with the original poster's confusion, clarifying the assumptions of the problem. Some guidance has been offered regarding the interpretation of the function's properties, and the original poster acknowledges a misunderstanding after receiving feedback.

Contextual Notes

The original poster's example was intended to illustrate a point but inadvertently contradicted the problem's conditions. There is a focus on ensuring clarity regarding the definitions and assumptions involved in the discussion.

Portuga
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Consider a continuous function f in [a,b] and f(a) < f(b). Suppose that \forall s \neq t in [a,b], f(s) \neq f(t). Proof that f is strictly increasing function in [a,b].

Homework Equations



I.V.T: If f is continuous in [a,b] and \gamma is a real in [f(a),f(b)], then there'll be at least one c in [a,b] such that f(c) = \gamma.

The Attempt at a Solution



This exercise is very strange to me. Besides I can apply the I.V.T to show that for any sub interval in [a,b] there will be an intermediate value in f(a), f(b), I can easily draw and counter example of what it pretends:
https://www.dropbox.com/s/dtj28xo4ilaai4z/pf.eps?dl=0

I am missing something important?

Thanks in advance!
 
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Portuga said:
Consider a continuous function f in [a,b] and f(a) < f(b). Suppose that \forall s \neq t in [a,b], f(s) \neq f(t). Proof that f is strictly increasing function in [a,b].

Homework Equations



I.V.T: If f is continuous in [a,b] and \gamma is a real in [f(a),f(b)], then there'll be at least one c in [a,b] such that f(c) = \gamma.

The Attempt at a Solution



This exercise is very strange to me. Besides I can apply the I.V.T to show that for any sub interval in [a,b] there will be an intermediate value in f(a), f(b), I can easily draw and counter example of what it pretends:
https://www.dropbox.com/s/dtj28xo4ilaai4z/pf.eps?dl=0

I am missing something important?

Thanks in advance!

Your example violates the hypotheses of the claim: your function ##f(x)## has ##f(s) = f(t)## for several pairs ##(s,t)## with ##s \neq t##.
 
Picture doesn't make it. Blank screen. Can you copy/paste it in the post ?
 
BvU said:
Picture doesn't make it. Blank screen. Can you copy/paste it in the post ?
upload_2017-7-5_11-24-35.png
 
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I did the first one in libreoffice, so I made the mistake pointed by Ray.
 
Portuga said:
Portuga said:
Suppose that \forall s \neq t in [a,b], f(s) \neq f(t).
Your drawing violates the assumption that ##f(s) \neq f(t)##
Portuga said:
Proof that f is strictly increasing function in [a,b].

Minor point. The verb is "to prove". The noun is "proof".
 
Thank you! Now I got the point! Sorry for my poor English! Thank you very much. Now it's clear for me!
 
Portuga said:
Sorry for my poor English!
No need for an apology. Lots of native speakers of English also get this wrong (prove vs. proof), sometimes spelling "prove" as "proove."
 
Figure from link in OP:
upload_2017-7-8_11-40-23.png
 

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