How to approach this functional equation question?












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Given that $forall x,y in mathbb{R}, f(x+y)=f(x)+xcdot y + y$, determine $f'(x)$.



I tried plugging in spacial cases like letting $x=y=0$, but it did not seem to get me anywhere.



(The question does not state anything about continuity. What should I assume in such cases?)






functional-equations







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    3












    $begingroup$


    Given that $forall x,y in mathbb{R}, f(x+y)=f(x)+xcdot y + y$, determine $f'(x)$.



    I tried plugging in spacial cases like letting $x=y=0$, but it did not seem to get me anywhere.



    (The question does not state anything about continuity. What should I assume in such cases?)






    functional-equations







    share|cite|improve this question









    New contributor




    joeblack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3





      $begingroup$


      Given that $forall x,y in mathbb{R}, f(x+y)=f(x)+xcdot y + y$, determine $f'(x)$.



      I tried plugging in spacial cases like letting $x=y=0$, but it did not seem to get me anywhere.



      (The question does not state anything about continuity. What should I assume in such cases?)






      functional-equations







      share|cite|improve this question









      New contributor




      joeblack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Given that $forall x,y in mathbb{R}, f(x+y)=f(x)+xcdot y + y$, determine $f'(x)$.



      I tried plugging in spacial cases like letting $x=y=0$, but it did not seem to get me anywhere.



      (The question does not state anything about continuity. What should I assume in such cases?)






      functional-equations




      functional-equations






      share|cite|improve this question









      New contributor




      joeblack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question




      share|cite|improve this question








      edited 29 mins ago







      joeblack













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      asked 34 mins ago









      joeblackjoeblack

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          3 Answers
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          active

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          5












          $begingroup$

          Answer: $f'$ does not exists.



          If we put $y=-x$ we get $$f(0) = f(x)-x^2-x$$ so $$f(x)=x^2+x+a$$ where $a=f(0)$ So Pluggin this in to the starting equation we get $$(x+y)^2+x+y+a = x^2+x+a+xy+y$$



          and we get $$y(x+y) =0;;;forall x,y$$



          A contradiction. Such a function does not exists, so $f'(x)$ does not also exists.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
            $endgroup$
            – Sil
            19 mins ago





















          1












          $begingroup$

          Hint $:$ For all $h neq 0$ observe that $frac {f(x+h) - f(x)} {h} = x + 1.$ So $$f'(x)=limlimits_{h rightarrow 0} frac {f(x+h) - f(x)} {h} = x+1.$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Regarding continuity, we could use the given functional equation to prove that $f$ must be continuous everywhere. First of all, it should be clear that $f$ is defined on the whole real line. Such a function $f$ will be continuous if and only if for all $epsilon>0$ and all $xinBbb R,$ there is some $delta>0$ such that $$bigl|f(x+y)-f(x)bigr|<epsilon$$ for all $yin(-delta,delta).$ From the functional equation, this means we must show that for all $epsilon>0$ and $xinBbb R,$ there is some $delta>0$ such that $|x+1||y|<epsilon$ whenever $|y|<delta.$ Would you be able to take it from there?



            However, it would be even more straightforward to prove that $f$ is differentiable everywhere (by explicitly calculating the derivative), since differentiable functions are continuous. begin{eqnarray}f'(x) &:=& lim_{hto 0}frac{f(x+h)-f(x)}h\ &=& lim_{hto 0}frac{f(x)+xcdot h+h-f(x)}h\ &=& lim_{hto 0}frac{(x+1)cdot h}h\ &=& lim_{hto 0}x+1\ &=& x+1.end{eqnarray}



            As a result, we can say that $f(x)=frac12x^2+x+c$ for some real $c.$ Our job, then, is to determine what value(s) of $c$ (if any) will allow $f$ to satisfy the given functional equation.



            Well, by the functional equation, we can see that we need begin{eqnarray}f(x)+xcdot y+y &=& f(x+y)\ &=& frac12(x+y)^2+(x+y)+c\ &=& frac12left(x^2+2xcdot y+y^2right)+x+y+c\ &=& frac12x^2+xcdot y+frac12y^2+x+y+c\ &=& f(x)+xcdot y+y+frac12y^2end{eqnarray} for all real $x,y.$ Unfortunately, this only holds when $y=0,$ so no value of $c$ will work. Thus, $f$ does not exist, nor does $f'$.






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              3 Answers
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              3 Answers
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              5












              $begingroup$

              Answer: $f'$ does not exists.



              If we put $y=-x$ we get $$f(0) = f(x)-x^2-x$$ so $$f(x)=x^2+x+a$$ where $a=f(0)$ So Pluggin this in to the starting equation we get $$(x+y)^2+x+y+a = x^2+x+a+xy+y$$



              and we get $$y(x+y) =0;;;forall x,y$$



              A contradiction. Such a function does not exists, so $f'(x)$ does not also exists.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
                $endgroup$
                – Sil
                19 mins ago


















              5












              $begingroup$

              Answer: $f'$ does not exists.



              If we put $y=-x$ we get $$f(0) = f(x)-x^2-x$$ so $$f(x)=x^2+x+a$$ where $a=f(0)$ So Pluggin this in to the starting equation we get $$(x+y)^2+x+y+a = x^2+x+a+xy+y$$



              and we get $$y(x+y) =0;;;forall x,y$$



              A contradiction. Such a function does not exists, so $f'(x)$ does not also exists.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
                $endgroup$
                – Sil
                19 mins ago
















              5












              5








              5





              $begingroup$

              Answer: $f'$ does not exists.



              If we put $y=-x$ we get $$f(0) = f(x)-x^2-x$$ so $$f(x)=x^2+x+a$$ where $a=f(0)$ So Pluggin this in to the starting equation we get $$(x+y)^2+x+y+a = x^2+x+a+xy+y$$



              and we get $$y(x+y) =0;;;forall x,y$$



              A contradiction. Such a function does not exists, so $f'(x)$ does not also exists.






              share|cite|improve this answer











              $endgroup$



              Answer: $f'$ does not exists.



              If we put $y=-x$ we get $$f(0) = f(x)-x^2-x$$ so $$f(x)=x^2+x+a$$ where $a=f(0)$ So Pluggin this in to the starting equation we get $$(x+y)^2+x+y+a = x^2+x+a+xy+y$$



              and we get $$y(x+y) =0;;;forall x,y$$



              A contradiction. Such a function does not exists, so $f'(x)$ does not also exists.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 22 mins ago

























              answered 27 mins ago









              Maria MazurMaria Mazur

              49.2k1360122




              49.2k1360122








              • 1




                $begingroup$
                $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
                $endgroup$
                – Sil
                19 mins ago
















              • 1




                $begingroup$
                $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
                $endgroup$
                – Sil
                19 mins ago










              1




              1




              $begingroup$
              $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
              $endgroup$
              – Sil
              19 mins ago






              $begingroup$
              $x=0$ also works, giving $f(y)=f(0)+y$ a linear function, another contradictory result...
              $endgroup$
              – Sil
              19 mins ago













              1












              $begingroup$

              Hint $:$ For all $h neq 0$ observe that $frac {f(x+h) - f(x)} {h} = x + 1.$ So $$f'(x)=limlimits_{h rightarrow 0} frac {f(x+h) - f(x)} {h} = x+1.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint $:$ For all $h neq 0$ observe that $frac {f(x+h) - f(x)} {h} = x + 1.$ So $$f'(x)=limlimits_{h rightarrow 0} frac {f(x+h) - f(x)} {h} = x+1.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint $:$ For all $h neq 0$ observe that $frac {f(x+h) - f(x)} {h} = x + 1.$ So $$f'(x)=limlimits_{h rightarrow 0} frac {f(x+h) - f(x)} {h} = x+1.$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint $:$ For all $h neq 0$ observe that $frac {f(x+h) - f(x)} {h} = x + 1.$ So $$f'(x)=limlimits_{h rightarrow 0} frac {f(x+h) - f(x)} {h} = x+1.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 29 mins ago









                  Dbchatto67Dbchatto67

                  2,437422




                  2,437422























                      1












                      $begingroup$

                      Regarding continuity, we could use the given functional equation to prove that $f$ must be continuous everywhere. First of all, it should be clear that $f$ is defined on the whole real line. Such a function $f$ will be continuous if and only if for all $epsilon>0$ and all $xinBbb R,$ there is some $delta>0$ such that $$bigl|f(x+y)-f(x)bigr|<epsilon$$ for all $yin(-delta,delta).$ From the functional equation, this means we must show that for all $epsilon>0$ and $xinBbb R,$ there is some $delta>0$ such that $|x+1||y|<epsilon$ whenever $|y|<delta.$ Would you be able to take it from there?



                      However, it would be even more straightforward to prove that $f$ is differentiable everywhere (by explicitly calculating the derivative), since differentiable functions are continuous. begin{eqnarray}f'(x) &:=& lim_{hto 0}frac{f(x+h)-f(x)}h\ &=& lim_{hto 0}frac{f(x)+xcdot h+h-f(x)}h\ &=& lim_{hto 0}frac{(x+1)cdot h}h\ &=& lim_{hto 0}x+1\ &=& x+1.end{eqnarray}



                      As a result, we can say that $f(x)=frac12x^2+x+c$ for some real $c.$ Our job, then, is to determine what value(s) of $c$ (if any) will allow $f$ to satisfy the given functional equation.



                      Well, by the functional equation, we can see that we need begin{eqnarray}f(x)+xcdot y+y &=& f(x+y)\ &=& frac12(x+y)^2+(x+y)+c\ &=& frac12left(x^2+2xcdot y+y^2right)+x+y+c\ &=& frac12x^2+xcdot y+frac12y^2+x+y+c\ &=& f(x)+xcdot y+y+frac12y^2end{eqnarray} for all real $x,y.$ Unfortunately, this only holds when $y=0,$ so no value of $c$ will work. Thus, $f$ does not exist, nor does $f'$.






                      share|cite









                      $endgroup$


















                        1












                        $begingroup$

                        Regarding continuity, we could use the given functional equation to prove that $f$ must be continuous everywhere. First of all, it should be clear that $f$ is defined on the whole real line. Such a function $f$ will be continuous if and only if for all $epsilon>0$ and all $xinBbb R,$ there is some $delta>0$ such that $$bigl|f(x+y)-f(x)bigr|<epsilon$$ for all $yin(-delta,delta).$ From the functional equation, this means we must show that for all $epsilon>0$ and $xinBbb R,$ there is some $delta>0$ such that $|x+1||y|<epsilon$ whenever $|y|<delta.$ Would you be able to take it from there?



                        However, it would be even more straightforward to prove that $f$ is differentiable everywhere (by explicitly calculating the derivative), since differentiable functions are continuous. begin{eqnarray}f'(x) &:=& lim_{hto 0}frac{f(x+h)-f(x)}h\ &=& lim_{hto 0}frac{f(x)+xcdot h+h-f(x)}h\ &=& lim_{hto 0}frac{(x+1)cdot h}h\ &=& lim_{hto 0}x+1\ &=& x+1.end{eqnarray}



                        As a result, we can say that $f(x)=frac12x^2+x+c$ for some real $c.$ Our job, then, is to determine what value(s) of $c$ (if any) will allow $f$ to satisfy the given functional equation.



                        Well, by the functional equation, we can see that we need begin{eqnarray}f(x)+xcdot y+y &=& f(x+y)\ &=& frac12(x+y)^2+(x+y)+c\ &=& frac12left(x^2+2xcdot y+y^2right)+x+y+c\ &=& frac12x^2+xcdot y+frac12y^2+x+y+c\ &=& f(x)+xcdot y+y+frac12y^2end{eqnarray} for all real $x,y.$ Unfortunately, this only holds when $y=0,$ so no value of $c$ will work. Thus, $f$ does not exist, nor does $f'$.






                        share|cite









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Regarding continuity, we could use the given functional equation to prove that $f$ must be continuous everywhere. First of all, it should be clear that $f$ is defined on the whole real line. Such a function $f$ will be continuous if and only if for all $epsilon>0$ and all $xinBbb R,$ there is some $delta>0$ such that $$bigl|f(x+y)-f(x)bigr|<epsilon$$ for all $yin(-delta,delta).$ From the functional equation, this means we must show that for all $epsilon>0$ and $xinBbb R,$ there is some $delta>0$ such that $|x+1||y|<epsilon$ whenever $|y|<delta.$ Would you be able to take it from there?



                          However, it would be even more straightforward to prove that $f$ is differentiable everywhere (by explicitly calculating the derivative), since differentiable functions are continuous. begin{eqnarray}f'(x) &:=& lim_{hto 0}frac{f(x+h)-f(x)}h\ &=& lim_{hto 0}frac{f(x)+xcdot h+h-f(x)}h\ &=& lim_{hto 0}frac{(x+1)cdot h}h\ &=& lim_{hto 0}x+1\ &=& x+1.end{eqnarray}



                          As a result, we can say that $f(x)=frac12x^2+x+c$ for some real $c.$ Our job, then, is to determine what value(s) of $c$ (if any) will allow $f$ to satisfy the given functional equation.



                          Well, by the functional equation, we can see that we need begin{eqnarray}f(x)+xcdot y+y &=& f(x+y)\ &=& frac12(x+y)^2+(x+y)+c\ &=& frac12left(x^2+2xcdot y+y^2right)+x+y+c\ &=& frac12x^2+xcdot y+frac12y^2+x+y+c\ &=& f(x)+xcdot y+y+frac12y^2end{eqnarray} for all real $x,y.$ Unfortunately, this only holds when $y=0,$ so no value of $c$ will work. Thus, $f$ does not exist, nor does $f'$.






                          share|cite









                          $endgroup$



                          Regarding continuity, we could use the given functional equation to prove that $f$ must be continuous everywhere. First of all, it should be clear that $f$ is defined on the whole real line. Such a function $f$ will be continuous if and only if for all $epsilon>0$ and all $xinBbb R,$ there is some $delta>0$ such that $$bigl|f(x+y)-f(x)bigr|<epsilon$$ for all $yin(-delta,delta).$ From the functional equation, this means we must show that for all $epsilon>0$ and $xinBbb R,$ there is some $delta>0$ such that $|x+1||y|<epsilon$ whenever $|y|<delta.$ Would you be able to take it from there?



                          However, it would be even more straightforward to prove that $f$ is differentiable everywhere (by explicitly calculating the derivative), since differentiable functions are continuous. begin{eqnarray}f'(x) &:=& lim_{hto 0}frac{f(x+h)-f(x)}h\ &=& lim_{hto 0}frac{f(x)+xcdot h+h-f(x)}h\ &=& lim_{hto 0}frac{(x+1)cdot h}h\ &=& lim_{hto 0}x+1\ &=& x+1.end{eqnarray}



                          As a result, we can say that $f(x)=frac12x^2+x+c$ for some real $c.$ Our job, then, is to determine what value(s) of $c$ (if any) will allow $f$ to satisfy the given functional equation.



                          Well, by the functional equation, we can see that we need begin{eqnarray}f(x)+xcdot y+y &=& f(x+y)\ &=& frac12(x+y)^2+(x+y)+c\ &=& frac12left(x^2+2xcdot y+y^2right)+x+y+c\ &=& frac12x^2+xcdot y+frac12y^2+x+y+c\ &=& f(x)+xcdot y+y+frac12y^2end{eqnarray} for all real $x,y.$ Unfortunately, this only holds when $y=0,$ so no value of $c$ will work. Thus, $f$ does not exist, nor does $f'$.







                          share|cite












                          share|cite



                          share|cite










                          answered 6 mins ago









                          Cameron BuieCameron Buie

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