Intermediate Value Theorem question

In summary, the conversation discusses the proof that a continuous function f in [a,b] with f(a) < f(b) and the property that for any s and t in [a,b], f(s) does not equal f(t), must be strictly increasing in [a,b]. The Intermediate Value Theorem is mentioned as a possible approach to proving this claim, but it is noted that a counterexample can be drawn for the given function.
  • #1
Portuga
56
6
Consider a continuous function [itex]f[/itex] in [itex][a,b][/itex] and [itex]f(a) < f(b)[/itex]. Suppose that [itex]\forall s \neq t[/itex] in [itex][a,b][/itex], [itex]f(s) \neq f(t)[/itex]. Proof that [itex]f[/itex] is strictly increasing function in [itex][a,b][/itex].

Homework Equations



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

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 [itex]f(a), f(b)[/itex], 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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  • #2
Portuga said:
Consider a continuous function [itex]f[/itex] in [itex][a,b][/itex] and [itex]f(a) < f(b)[/itex]. Suppose that [itex]\forall s \neq t[/itex] in [itex][a,b][/itex], [itex]f(s) \neq f(t)[/itex]. Proof that [itex]f[/itex] is strictly increasing function in [itex][a,b][/itex].

Homework Equations



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

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 [itex]f(a), f(b)[/itex], 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##.
 
  • #3
Picture doesn't make it. Blank screen. Can you copy/paste it in the post ?
 
  • #4
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
 
  • Like
Likes BvU
  • #5
I did the first one in libreoffice, so I made the mistake pointed by Ray.
 
  • #6
Portuga said:
View attachment 206618
Portuga said:
Suppose that [itex]\forall s \neq t[/itex] in [itex][a,b][/itex], [itex]f(s) \neq f(t)[/itex].
Your drawing violates the assumption that ##f(s) \neq f(t)##
Portuga said:
Proof that [itex]f[/itex] is strictly increasing function in [itex][a,b][/itex].

Minor point. The verb is "to prove". The noun is "proof".
 
  • #7
Thank you! Now I got the point! Sorry for my poor English! Thank you very much. Now it's clear for me!
 
  • #8
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."
 
  • #9
Figure from link in OP:
upload_2017-7-8_11-40-23.png
 

Related to Intermediate Value Theorem question

1. What is the Intermediate Value Theorem (IVT)?

The Intermediate Value Theorem is a mathematical theorem that states that if a continuous function f(x) is defined on the closed interval [a,b], and if y is any number between f(a) and f(b), then there exists at least one value c in [a,b] such that f(c) = y. In other words, the function takes on every intermediate value between its endpoints on the interval.

2. What is the significance of the Intermediate Value Theorem?

The IVT is significant because it provides a way to prove the existence of solutions for equations or inequalities involving continuous functions. It is commonly used in calculus to show the existence of roots of equations or to prove the existence of extrema.

3. How is the Intermediate Value Theorem used in real-world applications?

The IVT has many real-world applications, such as in economics, physics, and engineering. For example, it can be used to show the existence of a solution to a physical problem, such as finding the velocity of an object at a certain time, or to prove the existence of equilibrium points in economic models.

4. Can the Intermediate Value Theorem be extended to functions of more than one variable?

Yes, the Intermediate Value Theorem can be extended to functions of more than one variable. In this case, the theorem states that if a continuous function f(x,y) is defined on a closed region in the xy-plane, and if a point z is between f(a,b) and f(c,d), then there exists at least one point (x,y) in the region such that f(x,y) = z.

5. How does the Intermediate Value Theorem differ from the Mean Value Theorem?

The Intermediate Value Theorem and the Mean Value Theorem are both fundamental theorems in calculus, but they have different applications. The IVT is used to prove the existence of solutions for equations or inequalities, while the MVT is used to prove the existence of a specific value of a derivative. Additionally, the IVT only requires the function to be continuous, while the MVT requires the function to be differentiable in addition to being continuous.

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