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0

$begingroup$

If there exists a $K>0$ and for all $x,yinmathbbR$ it holds that $|f(x) - f(y)| leq Kcdot|y-x|^2$, then for all $xin mathbbR$ we have $f'(x)=0$.

I do not understand derivations so I absolutely don't know what to do with it :( I have to decide if it is true and prove it, or say it is false and give an example when it is not valid.

But I think that the letter $K$ has to do something with demarcation ("supra and from below")

real-analysis analysis derivatives proof-writing

share|cite|improve this question

edited Mar 15 at 16:51

Janka

asked Mar 15 at 16:11

JankaJanka

105

$endgroup$

closed as off-topic by Lee Mosher, Alex Provost, José Carlos Santos, jgon, Callus Mar 15 at 21:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee Mosher, jgon, Callus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you "do not understand derivations" you cannot solve the problem. If you do "understand derivations", or at least the definition of the derivative, the solution is obvious.
  $endgroup$
  – Christian Blatter
  Mar 15 at 16:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you help me somehow, if the solution is obvious ?
  $endgroup$
  – Janka
  Mar 15 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 15 at 19:12
  

  
  
  
  

add a comment | 

0

$begingroup$

If there exists a $K>0$ and for all $x,yinmathbbR$ it holds that $|f(x) - f(y)| leq Kcdot|y-x|^2$, then for all $xin mathbbR$ we have $f'(x)=0$.

I do not understand derivations so I absolutely don't know what to do with it :( I have to decide if it is true and prove it, or say it is false and give an example when it is not valid.

But I think that the letter $K$ has to do something with demarcation ("supra and from below")

real-analysis analysis derivatives proof-writing

share|cite|improve this question

edited Mar 15 at 16:51

Janka

asked Mar 15 at 16:11

JankaJanka

105

$endgroup$

closed as off-topic by Lee Mosher, Alex Provost, José Carlos Santos, jgon, Callus Mar 15 at 21:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee Mosher, jgon, Callus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you "do not understand derivations" you cannot solve the problem. If you do "understand derivations", or at least the definition of the derivative, the solution is obvious.
  $endgroup$
  – Christian Blatter
  Mar 15 at 16:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you help me somehow, if the solution is obvious ?
  $endgroup$
  – Janka
  Mar 15 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 15 at 19:12
  

  
  
  
  

add a comment | 

$begingroup$

If there exists a $K>0$ and for all $x,yinmathbbR$ it holds that $|f(x) - f(y)| leq Kcdot|y-x|^2$, then for all $xin mathbbR$ we have $f'(x)=0$.

I do not understand derivations so I absolutely don't know what to do with it :( I have to decide if it is true and prove it, or say it is false and give an example when it is not valid.

But I think that the letter $K$ has to do something with demarcation ("supra and from below")

real-analysis analysis derivatives proof-writing

share|cite|improve this question

edited Mar 15 at 16:51

Janka

asked Mar 15 at 16:11

JankaJanka

105

$endgroup$

share|cite|improve this question

edited Mar 15 at 16:51

Janka

asked Mar 15 at 16:11

JankaJanka

105

share|cite|improve this question

edited Mar 15 at 16:51

Janka

asked Mar 15 at 16:11

JankaJanka

105

asked Mar 15 at 16:11

JankaJanka

105

asked Mar 15 at 16:11

JankaJanka

105

1 Answer
1

active

oldest

votes

1

$begingroup$

For any fixed $x_0inmathbb R$ the definition of the derivative $f'(x_0)$ is given by
$$f'(x_0)=lim_xto x_0f(x)-f(x_0)over x-x_0 ,$$
if this limit exists. Note that in the denominator we have $x-x_0 (ne0)$, and not $(x-x_0)^2$.

share|cite|improve this answer

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, but I didn't get it 😐
  $endgroup$
  – Janka
  Mar 15 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note $|f'(x_0)|=lim_xrightarrow x_0fracleq lim_xrightarrow x_0Kfrac=lim_xrightarrow x_0K|x-x_0|=0$.
  $endgroup$
  – Floris Claassens
  Mar 15 at 17:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can be $|x-x_0|=0$ as you wrote ? It is the denominator of a fragment
  $endgroup$
  – Janka
  Mar 15 at 17:38
  

  
  
  
  

add a comment | 

1

$begingroup$

For any fixed $x_0inmathbb R$ the definition of the derivative $f'(x_0)$ is given by
$$f'(x_0)=lim_xto x_0f(x)-f(x_0)over x-x_0 ,$$
if this limit exists. Note that in the denominator we have $x-x_0 (ne0)$, and not $(x-x_0)^2$.

share|cite|improve this answer

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, but I didn't get it 😐
  $endgroup$
  – Janka
  Mar 15 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note $|f'(x_0)|=lim_xrightarrow x_0fracleq lim_xrightarrow x_0Kfrac=lim_xrightarrow x_0K|x-x_0|=0$.
  $endgroup$
  – Floris Claassens
  Mar 15 at 17:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can be $|x-x_0|=0$ as you wrote ? It is the denominator of a fragment
  $endgroup$
  – Janka
  Mar 15 at 17:38
  

  
  
  
  

add a comment | 

1

$begingroup$

For any fixed $x_0inmathbb R$ the definition of the derivative $f'(x_0)$ is given by
$$f'(x_0)=lim_xto x_0f(x)-f(x_0)over x-x_0 ,$$
if this limit exists. Note that in the denominator we have $x-x_0 (ne0)$, and not $(x-x_0)^2$.

share|cite|improve this answer

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, but I didn't get it 😐
  $endgroup$
  – Janka
  Mar 15 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note $|f'(x_0)|=lim_xrightarrow x_0fracleq lim_xrightarrow x_0Kfrac=lim_xrightarrow x_0K|x-x_0|=0$.
  $endgroup$
  – Floris Claassens
  Mar 15 at 17:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can be $|x-x_0|=0$ as you wrote ? It is the denominator of a fragment
  $endgroup$
  – Janka
  Mar 15 at 17:38
  

  
  
  
  

add a comment | 

$begingroup$

For any fixed $x_0inmathbb R$ the definition of the derivative $f'(x_0)$ is given by
$$f'(x_0)=lim_xto x_0f(x)-f(x_0)over x-x_0 ,$$
if this limit exists. Note that in the denominator we have $x-x_0 (ne0)$, and not $(x-x_0)^2$.

share|cite|improve this answer

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327

$endgroup$

share|cite|improve this answer

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327

answered Mar 15 at 16:58

Christian BlatterChristian Blatter

175k8115327