Does $lim_xto infty f(x)= infty$

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Let f be a twice differentiable function on R such that both $f'$ and $f''$
are strictly positive on R. Then $lim_xto infty f(x)= infty$.

I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.

If this result is true please exaplain why? I hope it won't be

real-analysis derivatives

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edited 2 days ago

G Tony Jacobs

25.5k43483

asked 2 days ago

Cloud JR

456311

add a comment | 

up vote
0
down vote

favorite

Let f be a twice differentiable function on R such that both $f'$ and $f''$
are strictly positive on R. Then $lim_xto infty f(x)= infty$.

I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.

If this result is true please exaplain why? I hope it won't be

real-analysis derivatives

share|cite|improve this question

edited 2 days ago

G Tony Jacobs

25.5k43483

asked 2 days ago

Cloud JR

456311

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let f be a twice differentiable function on R such that both $f'$ and $f''$
are strictly positive on R. Then $lim_xto infty f(x)= infty$.

I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.

If this result is true please exaplain why? I hope it won't be

real-analysis derivatives

share|cite|improve this question

edited 2 days ago

G Tony Jacobs

25.5k43483

asked 2 days ago

Cloud JR

456311

Let f be a twice differentiable function on R such that both $f'$ and $f''$
are strictly positive on R. Then $lim_xto infty f(x)= infty$.

I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.

If this result is true please exaplain why? I hope it won't be

real-analysis derivatives

share|cite|improve this question

edited 2 days ago

G Tony Jacobs

25.5k43483

asked 2 days ago

Cloud JR

456311

share|cite|improve this question

share|cite|improve this question

edited 2 days ago

G Tony Jacobs

25.5k43483

edited 2 days ago

G Tony Jacobs

25.5k43483

edited 2 days ago

G Tony Jacobs

25.5k43483

25.5k43483

asked 2 days ago

Cloud JR

456311

asked 2 days ago

Cloud JR

456311

asked 2 days ago

Cloud JR

456311

456311

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.

share|cite|improve this answer

answered 2 days ago

G Tony Jacobs

25.5k43483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
  – Cloud JR
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $f, f'$ are supposed to be striclty positive, not stricly increasing.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
  – G Tony Jacobs
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ups. Misread the problem. Sorry.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's the obvious counterexample once you read the question correctly.
  – wonko
  2 days ago
  

  
  
  
  

 | 
show 8 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.

share|cite|improve this answer

answered 2 days ago

G Tony Jacobs

25.5k43483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
  – Cloud JR
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $f, f'$ are supposed to be striclty positive, not stricly increasing.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
  – G Tony Jacobs
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ups. Misread the problem. Sorry.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's the obvious counterexample once you read the question correctly.
  – wonko
  2 days ago
  

  
  
  
  

 | 
show 8 more comments

up vote
2
down vote



accepted

Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.

share|cite|improve this answer

answered 2 days ago

G Tony Jacobs

25.5k43483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
  – Cloud JR
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $f, f'$ are supposed to be striclty positive, not stricly increasing.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
  – G Tony Jacobs
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ups. Misread the problem. Sorry.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's the obvious counterexample once you read the question correctly.
  – wonko
  2 days ago
  

  
  
  
  

 | 
show 8 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.

share|cite|improve this answer

answered 2 days ago

G Tony Jacobs

25.5k43483

Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.

share|cite|improve this answer

answered 2 days ago

G Tony Jacobs

25.5k43483

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

G Tony Jacobs

25.5k43483

answered 2 days ago

G Tony Jacobs

25.5k43483

answered 2 days ago

G Tony Jacobs

25.5k43483

25.5k43483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
  – Cloud JR
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $f, f'$ are supposed to be striclty positive, not stricly increasing.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
  – G Tony Jacobs
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ups. Misread the problem. Sorry.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's the obvious counterexample once you read the question correctly.
  – wonko
  2 days ago
  

  
  
  
  

 | 
show 8 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
  – Cloud JR
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $f, f'$ are supposed to be striclty positive, not stricly increasing.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
  – G Tony Jacobs
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ups. Misread the problem. Sorry.
  – Jonas
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's the obvious counterexample once you read the question correctly.
  – wonko
  2 days ago
  

  
  
  
  

But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
– Cloud JR
2 days ago

But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
– Cloud JR
2 days ago

$f, f'$ are supposed to be striclty positive, not stricly increasing.
– Jonas
2 days ago

$f, f'$ are supposed to be striclty positive, not stricly increasing.
– Jonas
2 days ago

@Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
– G Tony Jacobs
2 days ago

@Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
– G Tony Jacobs
2 days ago

1

1

Ups. Misread the problem. Sorry.
– Jonas
2 days ago

Ups. Misread the problem. Sorry.
– Jonas
2 days ago

1

1

It's the obvious counterexample once you read the question correctly.
– wonko
2 days ago

It's the obvious counterexample once you read the question correctly.
– wonko
2 days ago

 | 
show 8 more comments

 

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