Mfcttrf

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How do I evaluate this function for x=2015

Integers and integer functionsHow to determine if this function is one-to-oneFor which constants $a$ function is continuousIs it true that this function $f(n)=n^13$?A polynomial problem related to $P(x) = 5x^2015 + x^2 + x + 1$If $P(n)$ divides $P(P(n)-2015)$, prove that $P(-2015)=0$Unkown polynomial functionfind all continuous functions $f : [0,1] to mathbb Q$ such that $f(frac12)=frac20152016$Sums of $f(f(x))=1-x$$f(1)=1$, $f(x+5) ge f(x) + 5$, $f(x+1) le f(x) + 1$, and $g(x) = f(x) + 1 - x$. Which of the folowing statement are true?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

6

2

$begingroup$

Let $f$ be a continuous function defined over the real numbers. Given that
$f(2017)=2016$ and
$$f(x)cdot (f(f(x))=1$$ for all real $x$.
Find $f(2015)$.

functions

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edited 8 hours ago

Aqua

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've changed the question to include it. Thanks!
  $endgroup$
  – NeedHelp
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So this function jumps from $1/2016$ to $2016$ and it is contiuous!?
  $endgroup$
  – Aqua
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Jumping from $frac 12016$ to $2016$ isn't weird. Afterall $f(x) = mx + b$ where $2016m + b = frac 12016$ and $2017m + b = 2016$ will do it. What is weird is that $f(x) = frac 1x$ for all $x: 1 le x le 2016$. The question is how does $f(x)$ behave on the interval $[2016, 2016 + epsilon)$. I have a feeling that this actually impossible
  $endgroup$
  – fleablood
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd actually be really interested if finding the function itself is possible
  $endgroup$
  – NeedHelp
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm pretty sure such a function is impossible. The must be a supremum $k: 2016le k < 2017$ where $f(x) = frac 1x$ but we can make the argument for all $x: k < x < 2017$ as well so $k ge 2017$ which is not possible a
  $endgroup$
  – fleablood
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

6

2

$begingroup$

Let $f$ be a continuous function defined over the real numbers. Given that
$f(2017)=2016$ and
$$f(x)cdot (f(f(x))=1$$ for all real $x$.
Find $f(2015)$.

functions

share|cite|improve this question

edited 8 hours ago

Aqua

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asked 8 hours ago

NeedHelpNeedHelp

3655 bronze badges

New contributor

NeedHelp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've changed the question to include it. Thanks!
  $endgroup$
  – NeedHelp
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So this function jumps from $1/2016$ to $2016$ and it is contiuous!?
  $endgroup$
  – Aqua
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Jumping from $frac 12016$ to $2016$ isn't weird. Afterall $f(x) = mx + b$ where $2016m + b = frac 12016$ and $2017m + b = 2016$ will do it. What is weird is that $f(x) = frac 1x$ for all $x: 1 le x le 2016$. The question is how does $f(x)$ behave on the interval $[2016, 2016 + epsilon)$. I have a feeling that this actually impossible
  $endgroup$
  – fleablood
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd actually be really interested if finding the function itself is possible
  $endgroup$
  – NeedHelp
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm pretty sure such a function is impossible. The must be a supremum $k: 2016le k < 2017$ where $f(x) = frac 1x$ but we can make the argument for all $x: k < x < 2017$ as well so $k ge 2017$ which is not possible a
  $endgroup$
  – fleablood
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let $f$ be a continuous function defined over the real numbers. Given that
$f(2017)=2016$ and
$$f(x)cdot (f(f(x))=1$$ for all real $x$.
Find $f(2015)$.

functions

share|cite|improve this question

edited 8 hours ago

Aqua

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NeedHelpNeedHelp

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$endgroup$

share|cite|improve this question

edited 8 hours ago

Aqua

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asked 8 hours ago

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share|cite|improve this question

edited 8 hours ago

Aqua

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edited 8 hours ago

Aqua

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edited 8 hours ago

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asked 8 hours ago

NeedHelpNeedHelp

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3 Answers
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13

$begingroup$

We have $1 = f(2017)cdot f(f(2017)) = 2016cdot f(2016)$, so $f(2016) = 1/2016$.

Since $f$ is continuous and $f(2016) = 1/2016 < 2015 < 2016 = f(2017)$, the intermediate value theorem tells us there is some $x$ between $2016$ and $2017$ such that $f(x) = 2015$.

Now we know $1 = f(x)cdot f(f(x)) = 2015cdot f(2015)$, so $f(2015) = 1/2015$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2015 lies between 1/2016 and 2017
  $endgroup$
  – Satish Ramanathan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah got that sorry : )
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Let $R$ be the range of $f$. Since $f$ is continuous, this is an interval.
$y = f(x) in R$, your equation says $y f(y) = 1$, i.e. $f(y) = 1/y$. Thus $R$ must be an interval which is mapped into itself by the function $t to 1/t$. We know $2016 in R$,
so $1/2016 in R$, and since it's an interval $2015 in R$. Thus $f(2015) = 1/2015$.

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can we deduce what is $f$?
  $endgroup$
  – Aqua
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For $x$ outside of $R$, $f(x)$ can be anything that stays within $R$ and varies continuously with $x$.
  $endgroup$
  – Magma
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

fleablood tried to prove that such a function can't exist, but that's not true. The following function fullfills all conditions of the problem:

$$
f(x)=
begincases
2016 & text for x < frac12016,\
frac1x & text for frac12016 le x le 2016,\
left(2016-frac12016right)(x-2016)+frac12016 & text for 2016 < x le 2017,\
2016 & text for 2017 < x,\
endcases
$$

Each 'leg' of $f$ is continuous and since the legs have the same value/limit at the common 'joints', the whole of $f$ is continuous. The most complicated looking leg is the third, but it's just a linear function from point $(2016,frac12106)$ to point $(2017,2016)$.

The range of the first and fourth leg is just the value $2016$, the range of the second leg is the closed interval $[frac12016,2016]$, the range of the third leg is the half open interval $(frac12016,2016]$.

Taking this alltogether, we get that the range of $f$ is $[frac12016,2016]$.

That means $forall x in mathbb R: f(x) in [frac12016,2016]$, so by the definition of the second leg we have

$$forall x in mathbb R: f(f(x)) = frac1f(x),$$

from which the functional equation of the problem follows.

share|cite|improve this answer

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges

$endgroup$

add a comment | 

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13

$begingroup$

We have $1 = f(2017)cdot f(f(2017)) = 2016cdot f(2016)$, so $f(2016) = 1/2016$.

Since $f$ is continuous and $f(2016) = 1/2016 < 2015 < 2016 = f(2017)$, the intermediate value theorem tells us there is some $x$ between $2016$ and $2017$ such that $f(x) = 2015$.

Now we know $1 = f(x)cdot f(f(x)) = 2015cdot f(2015)$, so $f(2015) = 1/2015$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2015 lies between 1/2016 and 2017
  $endgroup$
  – Satish Ramanathan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah got that sorry : )
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  

add a comment | 

13

$begingroup$

We have $1 = f(2017)cdot f(f(2017)) = 2016cdot f(2016)$, so $f(2016) = 1/2016$.

Since $f$ is continuous and $f(2016) = 1/2016 < 2015 < 2016 = f(2017)$, the intermediate value theorem tells us there is some $x$ between $2016$ and $2017$ such that $f(x) = 2015$.

Now we know $1 = f(x)cdot f(f(x)) = 2015cdot f(2015)$, so $f(2015) = 1/2015$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2015 lies between 1/2016 and 2017
  $endgroup$
  – Satish Ramanathan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah got that sorry : )
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We have $1 = f(2017)cdot f(f(2017)) = 2016cdot f(2016)$, so $f(2016) = 1/2016$.

Since $f$ is continuous and $f(2016) = 1/2016 < 2015 < 2016 = f(2017)$, the intermediate value theorem tells us there is some $x$ between $2016$ and $2017$ such that $f(x) = 2015$.

Now we know $1 = f(x)cdot f(f(x)) = 2015cdot f(2015)$, so $f(2015) = 1/2015$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

answered 8 hours ago

MagmaMagma

71211 silver badge88 bronze badges

6

$begingroup$

Let $R$ be the range of $f$. Since $f$ is continuous, this is an interval.
$y = f(x) in R$, your equation says $y f(y) = 1$, i.e. $f(y) = 1/y$. Thus $R$ must be an interval which is mapped into itself by the function $t to 1/t$. We know $2016 in R$,
so $1/2016 in R$, and since it's an interval $2015 in R$. Thus $f(2015) = 1/2015$.

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can we deduce what is $f$?
  $endgroup$
  – Aqua
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For $x$ outside of $R$, $f(x)$ can be anything that stays within $R$ and varies continuously with $x$.
  $endgroup$
  – Magma
  7 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Let $R$ be the range of $f$. Since $f$ is continuous, this is an interval.
$y = f(x) in R$, your equation says $y f(y) = 1$, i.e. $f(y) = 1/y$. Thus $R$ must be an interval which is mapped into itself by the function $t to 1/t$. We know $2016 in R$,
so $1/2016 in R$, and since it's an interval $2015 in R$. Thus $f(2015) = 1/2015$.

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can we deduce what is $f$?
  $endgroup$
  – Aqua
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For $x$ outside of $R$, $f(x)$ can be anything that stays within $R$ and varies continuously with $x$.
  $endgroup$
  – Magma
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $R$ be the range of $f$. Since $f$ is continuous, this is an interval.
$y = f(x) in R$, your equation says $y f(y) = 1$, i.e. $f(y) = 1/y$. Thus $R$ must be an interval which is mapped into itself by the function $t to 1/t$. We know $2016 in R$,
so $1/2016 in R$, and since it's an interval $2015 in R$. Thus $f(2015) = 1/2015$.

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

answered 8 hours ago

Robert IsraelRobert Israel

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1

$begingroup$

fleablood tried to prove that such a function can't exist, but that's not true. The following function fullfills all conditions of the problem:

$$
f(x)=
begincases
2016 & text for x < frac12016,\
frac1x & text for frac12016 le x le 2016,\
left(2016-frac12016right)(x-2016)+frac12016 & text for 2016 < x le 2017,\
2016 & text for 2017 < x,\
endcases
$$

Each 'leg' of $f$ is continuous and since the legs have the same value/limit at the common 'joints', the whole of $f$ is continuous. The most complicated looking leg is the third, but it's just a linear function from point $(2016,frac12106)$ to point $(2017,2016)$.

The range of the first and fourth leg is just the value $2016$, the range of the second leg is the closed interval $[frac12016,2016]$, the range of the third leg is the half open interval $(frac12016,2016]$.

Taking this alltogether, we get that the range of $f$ is $[frac12016,2016]$.

That means $forall x in mathbb R: f(x) in [frac12016,2016]$, so by the definition of the second leg we have

$$forall x in mathbb R: f(f(x)) = frac1f(x),$$

from which the functional equation of the problem follows.

share|cite|improve this answer

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

fleablood tried to prove that such a function can't exist, but that's not true. The following function fullfills all conditions of the problem:

$$
f(x)=
begincases
2016 & text for x < frac12016,\
frac1x & text for frac12016 le x le 2016,\
left(2016-frac12016right)(x-2016)+frac12016 & text for 2016 < x le 2017,\
2016 & text for 2017 < x,\
endcases
$$

Each 'leg' of $f$ is continuous and since the legs have the same value/limit at the common 'joints', the whole of $f$ is continuous. The most complicated looking leg is the third, but it's just a linear function from point $(2016,frac12106)$ to point $(2017,2016)$.

The range of the first and fourth leg is just the value $2016$, the range of the second leg is the closed interval $[frac12016,2016]$, the range of the third leg is the half open interval $(frac12016,2016]$.

Taking this alltogether, we get that the range of $f$ is $[frac12016,2016]$.

That means $forall x in mathbb R: f(x) in [frac12016,2016]$, so by the definition of the second leg we have

$$forall x in mathbb R: f(f(x)) = frac1f(x),$$

from which the functional equation of the problem follows.

share|cite|improve this answer

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges

$endgroup$

add a comment | 

$begingroup$

fleablood tried to prove that such a function can't exist, but that's not true. The following function fullfills all conditions of the problem:

$$
f(x)=
begincases
2016 & text for x < frac12016,\
frac1x & text for frac12016 le x le 2016,\
left(2016-frac12016right)(x-2016)+frac12016 & text for 2016 < x le 2017,\
2016 & text for 2017 < x,\
endcases
$$

Each 'leg' of $f$ is continuous and since the legs have the same value/limit at the common 'joints', the whole of $f$ is continuous. The most complicated looking leg is the third, but it's just a linear function from point $(2016,frac12106)$ to point $(2017,2016)$.

The range of the first and fourth leg is just the value $2016$, the range of the second leg is the closed interval $[frac12016,2016]$, the range of the third leg is the half open interval $(frac12016,2016]$.

Taking this alltogether, we get that the range of $f$ is $[frac12016,2016]$.

That means $forall x in mathbb R: f(x) in [frac12016,2016]$, so by the definition of the second leg we have

$$forall x in mathbb R: f(f(x)) = frac1f(x),$$

from which the functional equation of the problem follows.

share|cite|improve this answer

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges

$endgroup$

share|cite|improve this answer

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges

answered 6 hours ago

IngixIngix

7,27522 gold badges55 silver badges1010 bronze badges