Determinate if exist a function $f:Bbb R rightarrow Bbb R $ such that: $f(x+y)=max(xy,x)+min(xy,y)$

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Determinate if it exists a function $f:Bbb R rightarrow Bbb R $ such that:
$$f(x+y)=max(xy,x)+min(xy,y)$$

My try was to define $(x,y)=(2,2)$ because in this way $x+y=xy$, and it didn't work, but I don't know if this is enough to determinate that this function does not exist.

Any hints?

linear-algebra algebra-precalculus functions

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edited Aug 6 at 0:43

Cornman

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asked Aug 6 at 0:40

Rodrigo Pizarro

696117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $(x,y)=(2,2)$ we get $f(4)=8$ This is certainly not enough to show the function doesn't exit. Try some other values.
  – saulspatz
  Aug 6 at 0:52
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Determinate if it exists a function $f:Bbb R rightarrow Bbb R $ such that:
$$f(x+y)=max(xy,x)+min(xy,y)$$

My try was to define $(x,y)=(2,2)$ because in this way $x+y=xy$, and it didn't work, but I don't know if this is enough to determinate that this function does not exist.

Any hints?

linear-algebra algebra-precalculus functions

share|cite|improve this question

edited Aug 6 at 0:43

Cornman

2,60721128

asked Aug 6 at 0:40

Rodrigo Pizarro

696117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $(x,y)=(2,2)$ we get $f(4)=8$ This is certainly not enough to show the function doesn't exit. Try some other values.
  – saulspatz
  Aug 6 at 0:52
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Determinate if it exists a function $f:Bbb R rightarrow Bbb R $ such that:
$$f(x+y)=max(xy,x)+min(xy,y)$$

My try was to define $(x,y)=(2,2)$ because in this way $x+y=xy$, and it didn't work, but I don't know if this is enough to determinate that this function does not exist.

Any hints?

linear-algebra algebra-precalculus functions

share|cite|improve this question

edited Aug 6 at 0:43

Cornman

2,60721128

asked Aug 6 at 0:40

Rodrigo Pizarro

696117

Determinate if it exists a function $f:Bbb R rightarrow Bbb R $ such that:
$$f(x+y)=max(xy,x)+min(xy,y)$$

My try was to define $(x,y)=(2,2)$ because in this way $x+y=xy$, and it didn't work, but I don't know if this is enough to determinate that this function does not exist.

Any hints?

linear-algebra algebra-precalculus functions

share|cite|improve this question

edited Aug 6 at 0:43

Cornman

2,60721128

asked Aug 6 at 0:40

Rodrigo Pizarro

696117

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 0:43

Cornman

2,60721128

edited Aug 6 at 0:43

Cornman

2,60721128

edited Aug 6 at 0:43

Cornman

2,60721128

2,60721128

asked Aug 6 at 0:40

Rodrigo Pizarro

696117

asked Aug 6 at 0:40

Rodrigo Pizarro

696117

asked Aug 6 at 0:40

Rodrigo Pizarro

696117

696117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $(x,y)=(2,2)$ we get $f(4)=8$ This is certainly not enough to show the function doesn't exit. Try some other values.
  – saulspatz
  Aug 6 at 0:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $(x,y)=(2,2)$ we get $f(4)=8$ This is certainly not enough to show the function doesn't exit. Try some other values.
  – saulspatz
  Aug 6 at 0:52
  

  
  
  
  

If $(x,y)=(2,2)$ we get $f(4)=8$ This is certainly not enough to show the function doesn't exit. Try some other values.
– saulspatz
Aug 6 at 0:52

If $(x,y)=(2,2)$ we get $f(4)=8$ This is certainly not enough to show the function doesn't exit. Try some other values.
– saulspatz
Aug 6 at 0:52

add a comment | 

2 Answers
2

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2
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accepted

No, such function can not exist, because it would not be well-defined.
For $rinmathbbR$ the function $f$ could possibly take infinite solutions, since you can write $r=x+y$ in an infinite amout of ways.

For example: $1=frac12+frac12$ and $1=2-1$.

Then $f(frac12+frac12)=max(frac14,frac12)+min(frac14,frac12)=frac12+frac14=frac34$

Otherwise:

$f(2-1)=max(-2,2)+min(-2,-1)=2+(-2)=0$

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answered Aug 6 at 1:00

Cornman

2,60721128

add a comment | 

up vote
3
down vote

HINT: Let $x=1$ and $y=2$. Then let $x=2$ and $y=1$.

share|cite|improve this answer

answered Aug 6 at 0:50

Frpzzd

17k63490

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

No, such function can not exist, because it would not be well-defined.
For $rinmathbbR$ the function $f$ could possibly take infinite solutions, since you can write $r=x+y$ in an infinite amout of ways.

For example: $1=frac12+frac12$ and $1=2-1$.

Then $f(frac12+frac12)=max(frac14,frac12)+min(frac14,frac12)=frac12+frac14=frac34$

Otherwise:

$f(2-1)=max(-2,2)+min(-2,-1)=2+(-2)=0$

share|cite|improve this answer

answered Aug 6 at 1:00

Cornman

2,60721128

add a comment | 

up vote
2
down vote



accepted

No, such function can not exist, because it would not be well-defined.
For $rinmathbbR$ the function $f$ could possibly take infinite solutions, since you can write $r=x+y$ in an infinite amout of ways.

For example: $1=frac12+frac12$ and $1=2-1$.

Then $f(frac12+frac12)=max(frac14,frac12)+min(frac14,frac12)=frac12+frac14=frac34$

Otherwise:

$f(2-1)=max(-2,2)+min(-2,-1)=2+(-2)=0$

share|cite|improve this answer

answered Aug 6 at 1:00

Cornman

2,60721128

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

No, such function can not exist, because it would not be well-defined.
For $rinmathbbR$ the function $f$ could possibly take infinite solutions, since you can write $r=x+y$ in an infinite amout of ways.

For example: $1=frac12+frac12$ and $1=2-1$.

Then $f(frac12+frac12)=max(frac14,frac12)+min(frac14,frac12)=frac12+frac14=frac34$

Otherwise:

$f(2-1)=max(-2,2)+min(-2,-1)=2+(-2)=0$

share|cite|improve this answer

answered Aug 6 at 1:00

Cornman

2,60721128

No, such function can not exist, because it would not be well-defined.
For $rinmathbbR$ the function $f$ could possibly take infinite solutions, since you can write $r=x+y$ in an infinite amout of ways.

For example: $1=frac12+frac12$ and $1=2-1$.

Then $f(frac12+frac12)=max(frac14,frac12)+min(frac14,frac12)=frac12+frac14=frac34$

Otherwise:

$f(2-1)=max(-2,2)+min(-2,-1)=2+(-2)=0$

share|cite|improve this answer

answered Aug 6 at 1:00

Cornman

2,60721128

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 1:00

Cornman

2,60721128

answered Aug 6 at 1:00

Cornman

2,60721128

answered Aug 6 at 1:00

Cornman

2,60721128

2,60721128

add a comment | 

add a comment | 

up vote
3
down vote

HINT: Let $x=1$ and $y=2$. Then let $x=2$ and $y=1$.

share|cite|improve this answer

answered Aug 6 at 0:50

Frpzzd

17k63490

add a comment | 

up vote
3
down vote

HINT: Let $x=1$ and $y=2$. Then let $x=2$ and $y=1$.

share|cite|improve this answer

answered Aug 6 at 0:50

Frpzzd

17k63490

add a comment | 

up vote
3
down vote

up vote
3
down vote

HINT: Let $x=1$ and $y=2$. Then let $x=2$ and $y=1$.

share|cite|improve this answer

answered Aug 6 at 0:50

Frpzzd

17k63490

HINT: Let $x=1$ and $y=2$. Then let $x=2$ and $y=1$.

share|cite|improve this answer

answered Aug 6 at 0:50

Frpzzd

17k63490

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 0:50

Frpzzd

17k63490

answered Aug 6 at 0:50

Frpzzd

17k63490

answered Aug 6 at 0:50

Frpzzd

17k63490

17k63490

add a comment | 

add a comment | 

 

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