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5

2

$begingroup$

I'm studying Calculus and having a trouble solving this question.

1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?

2) What if $f(0)=0$, $f′(x)$ exists for all $x$?

I could easily find the counter example for 1) (Therefore it is false)

But I'm not sure about 2)

If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?

Please leave a comment if you don't mind :)

calculus derivatives

share|cite|improve this question

edited 2 days ago

user21820

39.6k543156

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

New contributor

Mighty QWERTY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
  $endgroup$
  – b00n heT
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
  $endgroup$
  – littleO
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Don't remove relevant information from your question!
  $endgroup$
  – user21820
  2 days ago
  

  
  
  
  

add a comment | 

5

2

$begingroup$

I'm studying Calculus and having a trouble solving this question.

1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?

2) What if $f(0)=0$, $f′(x)$ exists for all $x$?

I could easily find the counter example for 1) (Therefore it is false)

But I'm not sure about 2)

If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?

Please leave a comment if you don't mind :)

calculus derivatives

share|cite|improve this question

edited 2 days ago

user21820

39.6k543156

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

New contributor

Mighty QWERTY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
  $endgroup$
  – b00n heT
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
  $endgroup$
  – littleO
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Don't remove relevant information from your question!
  $endgroup$
  – user21820
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm studying Calculus and having a trouble solving this question.

1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?

2) What if $f(0)=0$, $f′(x)$ exists for all $x$?

I could easily find the counter example for 1) (Therefore it is false)

But I'm not sure about 2)

If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?

Please leave a comment if you don't mind :)

calculus derivatives

share|cite|improve this question

edited 2 days ago

user21820

39.6k543156

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

New contributor

Mighty QWERTY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 days ago

user21820

39.6k543156

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

New contributor

Mighty QWERTY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 days ago

user21820

39.6k543156

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

New contributor

Mighty QWERTY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

user21820

39.6k543156

edited 2 days ago

user21820

39.6k543156

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

New contributor

Mighty QWERTY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

Mighty QWERTYMighty QWERTY

325

3 Answers
3

active

oldest

votes

4

$begingroup$

Hint: Consider $$f(x)=x-A sin^2 x $$
for large $A$.

share|cite|improve this answer

answered 2 days ago

user1337user1337

16.8k43592

$endgroup$

add a comment | 

4

$begingroup$

Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^-x$ has $f'(x)gt1$ for all $x$, while $f(x)=x-1over2x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.

Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^-x$, of course, does not satisfy that condition.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

$endgroup$

add a comment | 

-2

$begingroup$

Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.

share|cite|improve this answer

edited 2 days ago

Max

9011318

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11

New contributor

Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)leq x$ fails for $x$ negative.
  $endgroup$
  – Wojowu
  2 days ago
  

  
  
  
  

add a comment | 

Your Answer

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4

$begingroup$

Hint: Consider $$f(x)=x-A sin^2 x $$
for large $A$.

share|cite|improve this answer

answered 2 days ago

user1337user1337

16.8k43592

$endgroup$

add a comment | 

4

$begingroup$

Hint: Consider $$f(x)=x-A sin^2 x $$
for large $A$.

share|cite|improve this answer

answered 2 days ago

user1337user1337

16.8k43592

$endgroup$

add a comment | 

$begingroup$

Hint: Consider $$f(x)=x-A sin^2 x $$
for large $A$.

share|cite|improve this answer

answered 2 days ago

user1337user1337

16.8k43592

$endgroup$

share|cite|improve this answer

answered 2 days ago

user1337user1337

16.8k43592

answered 2 days ago

user1337user1337

16.8k43592

answered 2 days ago

user1337user1337

16.8k43592

4

$begingroup$

Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^-x$ has $f'(x)gt1$ for all $x$, while $f(x)=x-1over2x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.

Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^-x$, of course, does not satisfy that condition.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

$endgroup$

add a comment | 

4

$begingroup$

Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^-x$ has $f'(x)gt1$ for all $x$, while $f(x)=x-1over2x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.

Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^-x$, of course, does not satisfy that condition.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

$endgroup$

add a comment | 

$begingroup$

Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^-x$ has $f'(x)gt1$ for all $x$, while $f(x)=x-1over2x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.

Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^-x$, of course, does not satisfy that condition.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

answered 2 days ago

Barry CipraBarry Cipra

60.3k654126

-2

$begingroup$

Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.

share|cite|improve this answer

edited 2 days ago

Max

9011318

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11

New contributor

Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)leq x$ fails for $x$ negative.
  $endgroup$
  – Wojowu
  2 days ago
  

  
  
  
  

add a comment | 

-2

$begingroup$

Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.

share|cite|improve this answer

edited 2 days ago

Max

9011318

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11

New contributor

Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)leq x$ fails for $x$ negative.
  $endgroup$
  – Wojowu
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.

share|cite|improve this answer

edited 2 days ago

Max

9011318

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11

New contributor

Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited 2 days ago

Max

9011318

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11

New contributor

Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

Max

9011318

edited 2 days ago

Max

9011318

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11

New contributor

Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

Alex KovalevskyAlex Kovalevsky

11