Multi-Choice Question: Differentiable function

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The discussion revolves around a differentiable function f defined on the interval [0,1] with f(0)=0 and f(1)=1. Participants analyze three statements regarding the function's properties, particularly focusing on the validity of statement b, which claims there exists a point c in [0,1] such that f'(c)=1. The Mean Value Theorem is referenced as a key tool for understanding the function's behavior, but some participants express difficulty in applying it correctly. Counter-examples are discussed to challenge the validity of the statements, particularly regarding the continuity of the derivative. Overall, the conversation emphasizes the importance of the Mean Value Theorem in determining the correct answer.
sergey_le
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Homework Statement
Let function ƒ be differentiable on interval [0, 1] and ƒ(0) = 0, ƒ(1) = 1.
Which of the following is true?
(Edited)
be f Differentiable function In section [0,1] and f(0)=0, f(1)=1. so:
a. f A monotonous function arises in section [0,1].
b. There is a point c∈[0,1] so that f'(c)=1.
c. There is a point c∈(0,1) where f has Local max.
I have to choose one correct answer.
 
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I don't see what you have done up until now in attempting to find the answer. A good start is to draw such a function, then think about things such as the mean value theorem and similar theorems. Then please post what you have and where you get stuck so that one may go on from there.
 
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such 1 Explains why a is wrong
such 2 Explains why b is wrong
I can't see why b is wrong, so I guess b is the correct answer.
But I can't prove it.
I tried with the Mean value theorem, and it didn't work.
I tried with Darboux's theorem but it didn't work either
 
I think you have a typo: you probably mean that your diagram (2) is a counter-example to (c).

Try looking at your two graphs; suppose someone tried to use one of them as a "counter-example" to b. Your drawn functions are continuously differentiable in that interval. I'll do (2); for (1), it's similar. At the minimum at x= m, you have f'(m)=0; you also have f'(1)>1. Since the function g(x)=f'(x) is continuous, there must be a k, m<k<1, so that g(k)= f'(k) = 1. So, that counter-example to (b) wouldn't work. I am not saying that this is a proof that (b) is correct, but the counter-counter proof might set you thinking.
 
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sergey_le said:
View attachment 254872
such 1 Explains why a is wrong
such 2 Explains why b is wrong
I can't see why b is wrong, so I guess b is the correct answer.
But I can't prove it.
I tried with the Mean value theorem, and it didn't work.
I tried with Darboux's theorem but it didn't work either
Why didn't the mean value theorem work?
 
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sergey_le said:
Homework Statement:: Differentiable function
Homework Equations:: Differentiable function

be f Differentiable function In section [0,1] and f(0)=0, f(1)=1. so:
a. f A monotonous function arises in section [0,1].
b. There is a point c∈[0,1] so that f'(c)=1.
c. There is a point c∈(0,1) where f has Local max.
I have to choose one correct answer.

I had difficulty with the statement of your problem.

It looks like it should be stated somewhat like the following:

Let function ƒ be differentiable on interval [0, 1] and ƒ(0) = 0, ƒ(1) = 1.​
Which of the following is true?​
a.​
b.​
c.​

Is this a correct statement of the problem?
 
SammyS said:
I had difficulty with the statement of your problem.

It looks like it should be stated somewhat like the following:

Let function ƒ be differentiable on interval [0, 1] and ƒ(0) = 0, ƒ(1) = 1.​
Which of the following is true?​
a.​
b.​
c.​

Is this a correct statement of the problem?
Yes.
My English is not good.
I try to translate my questions into English.
I'm sorry
 
nomadreid said:
I think you have a typo: you probably mean that your diagram (2) is a counter-example to (c).

Try looking at your two graphs; suppose someone tried to use one of them as a "counter-example" to b. Your drawn functions are continuously differentiable in that interval. I'll do (2); for (1), it's similar. At the minimum at x= m, you have f'(m)=0; you also have f'(1)>1. Since the function g(x)=f'(x) is continuous, there must be a k, m<k<1, so that g(k)= f'(k) = 1. So, that counter-example to (b) wouldn't work. I am not saying that this is a proof that (b) is correct, but the counter-counter proof might set you thinking.
First off I have a typo. Thanks for noticing. English is not good.
I didn't understand how you know that f'(1)>1 ?
Could you please explain to me?
 
sergey_le said:
First off I have a typo. Thanks for noticing. English is not good.
I didn't understand how you know that f'(1)>1 ?
Could you please explain to me?
I think you should look at the Mean Value Theorem.
 
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  • #10
PeroK said:
I think you should look at the Mean Value Theorem.
thank you friend . I'm just trying to figure out how to do it
 
  • #11
sergey_le said:
thank you friend . I'm just trying to figure out how to do it
What does the MVT (Mean Value Theorem) say?
 
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  • #12
sergey_le said:
I didn't understand how you know that f'(1)>1 ?
Could you please explain to me?
Note that the slope of the line between your endpoints is 1, and then look at the first diagram to the right in https://en.wikipedia.org/wiki/Mean_value_theorem, and then read the caption.
 
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  • #13
Thanks so much everyone.
I understand
 
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