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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  • 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














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





















  • 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












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













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












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 0:43









Cornman

2,60721128




2,60721128









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
















  • 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










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






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      2 Answers
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      2 Answers
      2






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      active

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      up vote
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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:



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






      share|cite|improve this answer

























        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























          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













          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













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 6 at 1:00









          Cornman

          2,60721128




          2,60721128




















              up vote
              3
              down vote













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






              share|cite|improve this answer

























                up vote
                3
                down vote













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






                share|cite|improve this answer























                  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













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







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 6 at 0:50









                  Frpzzd

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