If $2f(x-1)+f(1-x)=2x$ then $f(x)=?$












0












$begingroup$


We've been given a function:
$$2f(x-1)+f(1-x)=2x$$
So now someone please teach me how to find:
$$f(x)=???$$
Sorry! Really sorry but I don't have any ideas about it! Please tell me how can I solve this kind of questions?






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    I would start by substituting $x-1=y$. Then your functional equation becomes $$f(-y) = 2(y-1-f(y))$$
    $endgroup$
    – Crostul
    Dec 9 '18 at 20:28












  • $begingroup$
    What have you done?
    $endgroup$
    – greedoid
    Dec 16 '18 at 20:04










  • $begingroup$
    Ummm.. nothing yet!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:05










  • $begingroup$
    @greedoid I was exactly now thinkin' about it!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:06










  • $begingroup$
    @greedoid would you tell me what Dr.mathva has done?!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:08
















0












$begingroup$


We've been given a function:
$$2f(x-1)+f(1-x)=2x$$
So now someone please teach me how to find:
$$f(x)=???$$
Sorry! Really sorry but I don't have any ideas about it! Please tell me how can I solve this kind of questions?






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    I would start by substituting $x-1=y$. Then your functional equation becomes $$f(-y) = 2(y-1-f(y))$$
    $endgroup$
    – Crostul
    Dec 9 '18 at 20:28












  • $begingroup$
    What have you done?
    $endgroup$
    – greedoid
    Dec 16 '18 at 20:04










  • $begingroup$
    Ummm.. nothing yet!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:05










  • $begingroup$
    @greedoid I was exactly now thinkin' about it!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:06










  • $begingroup$
    @greedoid would you tell me what Dr.mathva has done?!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:08














0












0








0


1



$begingroup$


We've been given a function:
$$2f(x-1)+f(1-x)=2x$$
So now someone please teach me how to find:
$$f(x)=???$$
Sorry! Really sorry but I don't have any ideas about it! Please tell me how can I solve this kind of questions?






functions







share|cite|improve this question









$endgroup$




We've been given a function:
$$2f(x-1)+f(1-x)=2x$$
So now someone please teach me how to find:
$$f(x)=???$$
Sorry! Really sorry but I don't have any ideas about it! Please tell me how can I solve this kind of questions?






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 20:20









user602338user602338

1607




1607












  • $begingroup$
    I would start by substituting $x-1=y$. Then your functional equation becomes $$f(-y) = 2(y-1-f(y))$$
    $endgroup$
    – Crostul
    Dec 9 '18 at 20:28












  • $begingroup$
    What have you done?
    $endgroup$
    – greedoid
    Dec 16 '18 at 20:04










  • $begingroup$
    Ummm.. nothing yet!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:05










  • $begingroup$
    @greedoid I was exactly now thinkin' about it!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:06










  • $begingroup$
    @greedoid would you tell me what Dr.mathva has done?!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:08


















  • $begingroup$
    I would start by substituting $x-1=y$. Then your functional equation becomes $$f(-y) = 2(y-1-f(y))$$
    $endgroup$
    – Crostul
    Dec 9 '18 at 20:28












  • $begingroup$
    What have you done?
    $endgroup$
    – greedoid
    Dec 16 '18 at 20:04










  • $begingroup$
    Ummm.. nothing yet!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:05










  • $begingroup$
    @greedoid I was exactly now thinkin' about it!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:06










  • $begingroup$
    @greedoid would you tell me what Dr.mathva has done?!
    $endgroup$
    – user602338
    Dec 16 '18 at 20:08
















$begingroup$
I would start by substituting $x-1=y$. Then your functional equation becomes $$f(-y) = 2(y-1-f(y))$$
$endgroup$
– Crostul
Dec 9 '18 at 20:28






$begingroup$
I would start by substituting $x-1=y$. Then your functional equation becomes $$f(-y) = 2(y-1-f(y))$$
$endgroup$
– Crostul
Dec 9 '18 at 20:28














$begingroup$
What have you done?
$endgroup$
– greedoid
Dec 16 '18 at 20:04




$begingroup$
What have you done?
$endgroup$
– greedoid
Dec 16 '18 at 20:04












$begingroup$
Ummm.. nothing yet!
$endgroup$
– user602338
Dec 16 '18 at 20:05




$begingroup$
Ummm.. nothing yet!
$endgroup$
– user602338
Dec 16 '18 at 20:05












$begingroup$
@greedoid I was exactly now thinkin' about it!
$endgroup$
– user602338
Dec 16 '18 at 20:06




$begingroup$
@greedoid I was exactly now thinkin' about it!
$endgroup$
– user602338
Dec 16 '18 at 20:06












$begingroup$
@greedoid would you tell me what Dr.mathva has done?!
$endgroup$
– user602338
Dec 16 '18 at 20:08




$begingroup$
@greedoid would you tell me what Dr.mathva has done?!
$endgroup$
– user602338
Dec 16 '18 at 20:08










4 Answers
4






active

oldest

votes


















1












$begingroup$

$xto 1-x$: $$2f(-x)+f(x)=2-2x$$
$xto 1+x$: $$2f(x)+f(-x)=2+2x$$ $$Rightarrow 4f(x)+2f(-x)=4+4x$$



Subtracting: $$3f(x)=2+6x$$ $$Rightarrow f(x)=frac{2}{3}+2x$$






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    Let $t= x-1$ then we get $$2f(t)+f(-t)= 2t+2$$ (which is valid for all real $t$ since the linear function $xmapsto x-1$ is surjective) and if we swich $t$ with $-t$ we get



    $$2f(-t)+f(t)= -2t+2$$ So $$ 2(2t+2-2f(t))+f(t)= -2t+2$$



    which give us: $$f(t)= 2t+2/3$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
      $endgroup$
      – user602338
      Dec 9 '18 at 20:32












    • $begingroup$
      Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
      $endgroup$
      – greedoid
      Dec 9 '18 at 20:33






    • 1




      $begingroup$
      @greedoid $f(t)=2t+2/3$
      $endgroup$
      – Shubham Johri
      Dec 9 '18 at 20:42










    • $begingroup$
      No I mean how did you switch between $t$ and $-t$??
      $endgroup$
      – user602338
      Dec 9 '18 at 20:42










    • $begingroup$
      @ShubhamJohri Thanks
      $endgroup$
      – greedoid
      Dec 9 '18 at 20:44



















    2












    $begingroup$

    $2f(x-1)+f(1-x)=2xRightarrow 2f(y)+f(-y)=2(y+1) ,(1)Rightarrow 2f(-y)+f(y)=2(-y+1)Rightarrow -4f(-y)-2f(y)=-4(-y+1) , (2)$



    $(1)+(2)Rightarrow -3f(-y)=6y-2Rightarrow -3f(y)=-6y-2Rightarrow f(y)=2y+dfrac{2}{3}$






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      $2f(x-1)+f(1-x)=2x$



      Set $y=1-ximplies x=1-y$



      $2f(-y)+f(y)=2(1-y)$



      Set $y=x-1implies x=y+1$



      $2f(y)+f(-y)=2(1+y)$



      Eliminate $f(-y)$ between the two equations:



      $implies f(y)=2y+frac23$






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        $xto 1-x$: $$2f(-x)+f(x)=2-2x$$
        $xto 1+x$: $$2f(x)+f(-x)=2+2x$$ $$Rightarrow 4f(x)+2f(-x)=4+4x$$



        Subtracting: $$3f(x)=2+6x$$ $$Rightarrow f(x)=frac{2}{3}+2x$$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          $xto 1-x$: $$2f(-x)+f(x)=2-2x$$
          $xto 1+x$: $$2f(x)+f(-x)=2+2x$$ $$Rightarrow 4f(x)+2f(-x)=4+4x$$



          Subtracting: $$3f(x)=2+6x$$ $$Rightarrow f(x)=frac{2}{3}+2x$$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            $xto 1-x$: $$2f(-x)+f(x)=2-2x$$
            $xto 1+x$: $$2f(x)+f(-x)=2+2x$$ $$Rightarrow 4f(x)+2f(-x)=4+4x$$



            Subtracting: $$3f(x)=2+6x$$ $$Rightarrow f(x)=frac{2}{3}+2x$$






            share|cite|improve this answer











            $endgroup$



            $xto 1-x$: $$2f(-x)+f(x)=2-2x$$
            $xto 1+x$: $$2f(x)+f(-x)=2+2x$$ $$Rightarrow 4f(x)+2f(-x)=4+4x$$



            Subtracting: $$3f(x)=2+6x$$ $$Rightarrow f(x)=frac{2}{3}+2x$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 20:46

























            answered Dec 9 '18 at 20:30









            Dr. MathvaDr. Mathva

            1,096316




            1,096316























                6












                $begingroup$

                Let $t= x-1$ then we get $$2f(t)+f(-t)= 2t+2$$ (which is valid for all real $t$ since the linear function $xmapsto x-1$ is surjective) and if we swich $t$ with $-t$ we get



                $$2f(-t)+f(t)= -2t+2$$ So $$ 2(2t+2-2f(t))+f(t)= -2t+2$$



                which give us: $$f(t)= 2t+2/3$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:32












                • $begingroup$
                  Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:33






                • 1




                  $begingroup$
                  @greedoid $f(t)=2t+2/3$
                  $endgroup$
                  – Shubham Johri
                  Dec 9 '18 at 20:42










                • $begingroup$
                  No I mean how did you switch between $t$ and $-t$??
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:42










                • $begingroup$
                  @ShubhamJohri Thanks
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:44
















                6












                $begingroup$

                Let $t= x-1$ then we get $$2f(t)+f(-t)= 2t+2$$ (which is valid for all real $t$ since the linear function $xmapsto x-1$ is surjective) and if we swich $t$ with $-t$ we get



                $$2f(-t)+f(t)= -2t+2$$ So $$ 2(2t+2-2f(t))+f(t)= -2t+2$$



                which give us: $$f(t)= 2t+2/3$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:32












                • $begingroup$
                  Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:33






                • 1




                  $begingroup$
                  @greedoid $f(t)=2t+2/3$
                  $endgroup$
                  – Shubham Johri
                  Dec 9 '18 at 20:42










                • $begingroup$
                  No I mean how did you switch between $t$ and $-t$??
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:42










                • $begingroup$
                  @ShubhamJohri Thanks
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:44














                6












                6








                6





                $begingroup$

                Let $t= x-1$ then we get $$2f(t)+f(-t)= 2t+2$$ (which is valid for all real $t$ since the linear function $xmapsto x-1$ is surjective) and if we swich $t$ with $-t$ we get



                $$2f(-t)+f(t)= -2t+2$$ So $$ 2(2t+2-2f(t))+f(t)= -2t+2$$



                which give us: $$f(t)= 2t+2/3$$






                share|cite|improve this answer











                $endgroup$



                Let $t= x-1$ then we get $$2f(t)+f(-t)= 2t+2$$ (which is valid for all real $t$ since the linear function $xmapsto x-1$ is surjective) and if we swich $t$ with $-t$ we get



                $$2f(-t)+f(t)= -2t+2$$ So $$ 2(2t+2-2f(t))+f(t)= -2t+2$$



                which give us: $$f(t)= 2t+2/3$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 9 '18 at 20:53

























                answered Dec 9 '18 at 20:29









                greedoidgreedoid

                40.7k1149100




                40.7k1149100












                • $begingroup$
                  Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:32












                • $begingroup$
                  Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:33






                • 1




                  $begingroup$
                  @greedoid $f(t)=2t+2/3$
                  $endgroup$
                  – Shubham Johri
                  Dec 9 '18 at 20:42










                • $begingroup$
                  No I mean how did you switch between $t$ and $-t$??
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:42










                • $begingroup$
                  @ShubhamJohri Thanks
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:44


















                • $begingroup$
                  Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:32












                • $begingroup$
                  Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:33






                • 1




                  $begingroup$
                  @greedoid $f(t)=2t+2/3$
                  $endgroup$
                  – Shubham Johri
                  Dec 9 '18 at 20:42










                • $begingroup$
                  No I mean how did you switch between $t$ and $-t$??
                  $endgroup$
                  – user602338
                  Dec 9 '18 at 20:42










                • $begingroup$
                  @ShubhamJohri Thanks
                  $endgroup$
                  – greedoid
                  Dec 9 '18 at 20:44
















                $begingroup$
                Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
                $endgroup$
                – user602338
                Dec 9 '18 at 20:32






                $begingroup$
                Oh big thanks to you. Just a little point: Can you explain how $f(-t)$ works?
                $endgroup$
                – user602338
                Dec 9 '18 at 20:32














                $begingroup$
                Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
                $endgroup$
                – greedoid
                Dec 9 '18 at 20:33




                $begingroup$
                Expres form 1.st $f(-t) = 2t+2-2f(t)$ and plug it in second equation.
                $endgroup$
                – greedoid
                Dec 9 '18 at 20:33




                1




                1




                $begingroup$
                @greedoid $f(t)=2t+2/3$
                $endgroup$
                – Shubham Johri
                Dec 9 '18 at 20:42




                $begingroup$
                @greedoid $f(t)=2t+2/3$
                $endgroup$
                – Shubham Johri
                Dec 9 '18 at 20:42












                $begingroup$
                No I mean how did you switch between $t$ and $-t$??
                $endgroup$
                – user602338
                Dec 9 '18 at 20:42




                $begingroup$
                No I mean how did you switch between $t$ and $-t$??
                $endgroup$
                – user602338
                Dec 9 '18 at 20:42












                $begingroup$
                @ShubhamJohri Thanks
                $endgroup$
                – greedoid
                Dec 9 '18 at 20:44




                $begingroup$
                @ShubhamJohri Thanks
                $endgroup$
                – greedoid
                Dec 9 '18 at 20:44











                2












                $begingroup$

                $2f(x-1)+f(1-x)=2xRightarrow 2f(y)+f(-y)=2(y+1) ,(1)Rightarrow 2f(-y)+f(y)=2(-y+1)Rightarrow -4f(-y)-2f(y)=-4(-y+1) , (2)$



                $(1)+(2)Rightarrow -3f(-y)=6y-2Rightarrow -3f(y)=-6y-2Rightarrow f(y)=2y+dfrac{2}{3}$






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  $2f(x-1)+f(1-x)=2xRightarrow 2f(y)+f(-y)=2(y+1) ,(1)Rightarrow 2f(-y)+f(y)=2(-y+1)Rightarrow -4f(-y)-2f(y)=-4(-y+1) , (2)$



                  $(1)+(2)Rightarrow -3f(-y)=6y-2Rightarrow -3f(y)=-6y-2Rightarrow f(y)=2y+dfrac{2}{3}$






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    $2f(x-1)+f(1-x)=2xRightarrow 2f(y)+f(-y)=2(y+1) ,(1)Rightarrow 2f(-y)+f(y)=2(-y+1)Rightarrow -4f(-y)-2f(y)=-4(-y+1) , (2)$



                    $(1)+(2)Rightarrow -3f(-y)=6y-2Rightarrow -3f(y)=-6y-2Rightarrow f(y)=2y+dfrac{2}{3}$






                    share|cite|improve this answer











                    $endgroup$



                    $2f(x-1)+f(1-x)=2xRightarrow 2f(y)+f(-y)=2(y+1) ,(1)Rightarrow 2f(-y)+f(y)=2(-y+1)Rightarrow -4f(-y)-2f(y)=-4(-y+1) , (2)$



                    $(1)+(2)Rightarrow -3f(-y)=6y-2Rightarrow -3f(y)=-6y-2Rightarrow f(y)=2y+dfrac{2}{3}$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 9 '18 at 20:48

























                    answered Dec 9 '18 at 20:28









                    giannispapavgiannispapav

                    1,534324




                    1,534324























                        1












                        $begingroup$

                        $2f(x-1)+f(1-x)=2x$



                        Set $y=1-ximplies x=1-y$



                        $2f(-y)+f(y)=2(1-y)$



                        Set $y=x-1implies x=y+1$



                        $2f(y)+f(-y)=2(1+y)$



                        Eliminate $f(-y)$ between the two equations:



                        $implies f(y)=2y+frac23$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          $2f(x-1)+f(1-x)=2x$



                          Set $y=1-ximplies x=1-y$



                          $2f(-y)+f(y)=2(1-y)$



                          Set $y=x-1implies x=y+1$



                          $2f(y)+f(-y)=2(1+y)$



                          Eliminate $f(-y)$ between the two equations:



                          $implies f(y)=2y+frac23$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $2f(x-1)+f(1-x)=2x$



                            Set $y=1-ximplies x=1-y$



                            $2f(-y)+f(y)=2(1-y)$



                            Set $y=x-1implies x=y+1$



                            $2f(y)+f(-y)=2(1+y)$



                            Eliminate $f(-y)$ between the two equations:



                            $implies f(y)=2y+frac23$






                            share|cite|improve this answer









                            $endgroup$



                            $2f(x-1)+f(1-x)=2x$



                            Set $y=1-ximplies x=1-y$



                            $2f(-y)+f(y)=2(1-y)$



                            Set $y=x-1implies x=y+1$



                            $2f(y)+f(-y)=2(1+y)$



                            Eliminate $f(-y)$ between the two equations:



                            $implies f(y)=2y+frac23$







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                            answered Dec 9 '18 at 20:34









                            Shubham JohriShubham Johri

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