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Using limit conditions to find a function

1

$begingroup$

I have a series of limits for a function and I was wondering what the best way would be to find the original function. I do know the answer but not sure what the best ways of going about finding the function is.

$$lim_xto 0f(x)= -infty$$ $$lim_xto ±inftyf(x)=0$$ $$lim_xto 3^-f(x)=infty$$ $$lim_xto 3^+f(x)=-infty$$

And finally $f(2)=0$. If it helps I can give the answer below but I shouldn't technically have it!

My first ideas were that the top must be $(2-x)$ as $f(2)=0$, but I am not sure about the denominator!

Thank you!

Hint: The function is $$f(x)=frac2-xx^2(x-3)$$

calculus limits functions

share|cite|improve this question

edited Apr 2 at 15:07

James odare

asked Apr 2 at 14:59

James odareJames odare

35613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have put the function in the question!
  $endgroup$
  – James odare
  Apr 2 at 15:07
  

  
  
  
  

add a comment | 

1

$begingroup$

I have a series of limits for a function and I was wondering what the best way would be to find the original function. I do know the answer but not sure what the best ways of going about finding the function is.

$$lim_xto 0f(x)= -infty$$ $$lim_xto ±inftyf(x)=0$$ $$lim_xto 3^-f(x)=infty$$ $$lim_xto 3^+f(x)=-infty$$

And finally $f(2)=0$. If it helps I can give the answer below but I shouldn't technically have it!

My first ideas were that the top must be $(2-x)$ as $f(2)=0$, but I am not sure about the denominator!

Thank you!

Hint: The function is $$f(x)=frac2-xx^2(x-3)$$

calculus limits functions

share|cite|improve this question

edited Apr 2 at 15:07

James odare

asked Apr 2 at 14:59

James odareJames odare

35613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have put the function in the question!
  $endgroup$
  – James odare
  Apr 2 at 15:07
  

  
  
  
  

add a comment | 

$begingroup$

I have a series of limits for a function and I was wondering what the best way would be to find the original function. I do know the answer but not sure what the best ways of going about finding the function is.

$$lim_xto 0f(x)= -infty$$ $$lim_xto ±inftyf(x)=0$$ $$lim_xto 3^-f(x)=infty$$ $$lim_xto 3^+f(x)=-infty$$

And finally $f(2)=0$. If it helps I can give the answer below but I shouldn't technically have it!

My first ideas were that the top must be $(2-x)$ as $f(2)=0$, but I am not sure about the denominator!

Thank you!

Hint: The function is $$f(x)=frac2-xx^2(x-3)$$

calculus limits functions

share|cite|improve this question

edited Apr 2 at 15:07

James odare

asked Apr 2 at 14:59

James odareJames odare

35613

$endgroup$

share|cite|improve this question

edited Apr 2 at 15:07

James odare

asked Apr 2 at 14:59

James odareJames odare

35613

share|cite|improve this question

edited Apr 2 at 15:07

James odare

asked Apr 2 at 14:59

James odareJames odare

35613

asked Apr 2 at 14:59

James odareJames odare

35613

asked Apr 2 at 14:59

James odareJames odare

35613

1 Answer
1

active

oldest

votes

2

$begingroup$

If you are after a rational function $dfracP(x)Q(x)$, then you need to have $deg P(x)<deg Q(x)$, so that $lim_xtopminftyf(x)=0$. Since $lim_xto0f(x)=-infty$ and $lim_xto3^-f(x)=infty$, $f$ will have to have a zero between $0$ and $3$ and you want to have $2$ as one such zero. And, since you want to have $lim_xto0f(x)=-infty$, it would be good if it behaved like $dfrac-1x^2$ there. Also, you want it to behave like $dfrac-1x-3$ near $3$. So, take$$f(x)=-fracx-2x^2(x-3).$$

share|cite|improve this answer

edited Apr 2 at 17:18

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This looks good, also $f(2)=0$ so maybe numerator should be $(2-x)$? Otherwise thank you!
  $endgroup$
  – James odare
  Apr 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot the condition $f(2)=0$! I shall edit my answer.
  $endgroup$
  – José Carlos Santos
  Apr 2 at 15:17
  

  
  
  
  

add a comment | 

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2

$begingroup$

If you are after a rational function $dfracP(x)Q(x)$, then you need to have $deg P(x)<deg Q(x)$, so that $lim_xtopminftyf(x)=0$. Since $lim_xto0f(x)=-infty$ and $lim_xto3^-f(x)=infty$, $f$ will have to have a zero between $0$ and $3$ and you want to have $2$ as one such zero. And, since you want to have $lim_xto0f(x)=-infty$, it would be good if it behaved like $dfrac-1x^2$ there. Also, you want it to behave like $dfrac-1x-3$ near $3$. So, take$$f(x)=-fracx-2x^2(x-3).$$

share|cite|improve this answer

edited Apr 2 at 17:18

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This looks good, also $f(2)=0$ so maybe numerator should be $(2-x)$? Otherwise thank you!
  $endgroup$
  – James odare
  Apr 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot the condition $f(2)=0$! I shall edit my answer.
  $endgroup$
  – José Carlos Santos
  Apr 2 at 15:17
  

  
  
  
  

add a comment | 

2

$begingroup$

If you are after a rational function $dfracP(x)Q(x)$, then you need to have $deg P(x)<deg Q(x)$, so that $lim_xtopminftyf(x)=0$. Since $lim_xto0f(x)=-infty$ and $lim_xto3^-f(x)=infty$, $f$ will have to have a zero between $0$ and $3$ and you want to have $2$ as one such zero. And, since you want to have $lim_xto0f(x)=-infty$, it would be good if it behaved like $dfrac-1x^2$ there. Also, you want it to behave like $dfrac-1x-3$ near $3$. So, take$$f(x)=-fracx-2x^2(x-3).$$

share|cite|improve this answer

edited Apr 2 at 17:18

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This looks good, also $f(2)=0$ so maybe numerator should be $(2-x)$? Otherwise thank you!
  $endgroup$
  – James odare
  Apr 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot the condition $f(2)=0$! I shall edit my answer.
  $endgroup$
  – José Carlos Santos
  Apr 2 at 15:17
  

  
  
  
  

add a comment | 

$begingroup$

If you are after a rational function $dfracP(x)Q(x)$, then you need to have $deg P(x)<deg Q(x)$, so that $lim_xtopminftyf(x)=0$. Since $lim_xto0f(x)=-infty$ and $lim_xto3^-f(x)=infty$, $f$ will have to have a zero between $0$ and $3$ and you want to have $2$ as one such zero. And, since you want to have $lim_xto0f(x)=-infty$, it would be good if it behaved like $dfrac-1x^2$ there. Also, you want it to behave like $dfrac-1x-3$ near $3$. So, take$$f(x)=-fracx-2x^2(x-3).$$

share|cite|improve this answer

edited Apr 2 at 17:18

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

share|cite|improve this answer

edited Apr 2 at 17:18

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247

answered Apr 2 at 15:11

José Carlos SantosJosé Carlos Santos

176k24137247