Real domain and range function to find all functions with nonzero x.












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Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.



My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.






algebra-precalculus functional-equations







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  • i did but it didn't really help
    – clache547
    Oct 14 '16 at 22:25
















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Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.



My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.






algebra-precalculus functional-equations







share|cite|improve this question
























  • i did but it didn't really help
    – clache547
    Oct 14 '16 at 22:25














0












0








0


1





Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.



My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.






algebra-precalculus functional-equations







share|cite|improve this question















Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.



My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.






algebra-precalculus functional-equations




algebra-precalculus functional-equations






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share|cite|improve this question













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edited Oct 14 '16 at 22:36









Simply Beautiful Art

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asked Oct 14 '16 at 22:16









clache547

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  • i did but it didn't really help
    – clache547
    Oct 14 '16 at 22:25


















  • i did but it didn't really help
    – clache547
    Oct 14 '16 at 22:25
















i did but it didn't really help
– clache547
Oct 14 '16 at 22:25




i did but it didn't really help
– clache547
Oct 14 '16 at 22:25










2 Answers
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3














$$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
$xtofrac{1}{1-x}$:
$$
fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
$$
$xto1-frac{1}{x}$:
$$
fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
$$
Hence, we have:
$$
begin{cases}
begin{align}
f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
end{align}
end{cases}
$$
Solve for $f(x)$:
$$
f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
$$






share|cite|improve this answer































    0














    Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$



    Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$



    So we have three equations to work with:
    begin{align}
    f(x)+ 3f(1-tfrac {1}{x})= 7x,\
    f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
    f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
    end{align}



    Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
    which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$



    Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$



    which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$



    Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
    $$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$



    for $x ne 0,1.$



    As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
    for $x=1.$



    (Also note that $f(x) = k$ for $x=0$)






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      active

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      3














      $$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
      $xtofrac{1}{1-x}$:
      $$
      fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
      $$
      $xto1-frac{1}{x}$:
      $$
      fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
      $$
      Hence, we have:
      $$
      begin{cases}
      begin{align}
      f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
      3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
      0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
      end{align}
      end{cases}
      $$
      Solve for $f(x)$:
      $$
      f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
      $$






      share|cite|improve this answer




























        3














        $$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
        $xtofrac{1}{1-x}$:
        $$
        fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
        $$
        $xto1-frac{1}{x}$:
        $$
        fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
        $$
        Hence, we have:
        $$
        begin{cases}
        begin{align}
        f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
        3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
        0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
        end{align}
        end{cases}
        $$
        Solve for $f(x)$:
        $$
        f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
        $$






        share|cite|improve this answer


























          3












          3








          3






          $$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
          $xtofrac{1}{1-x}$:
          $$
          fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
          $$
          $xto1-frac{1}{x}$:
          $$
          fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
          $$
          Hence, we have:
          $$
          begin{cases}
          begin{align}
          f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
          3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
          0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
          end{align}
          end{cases}
          $$
          Solve for $f(x)$:
          $$
          f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
          $$






          share|cite|improve this answer














          $$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
          $xtofrac{1}{1-x}$:
          $$
          fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
          $$
          $xto1-frac{1}{x}$:
          $$
          fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
          $$
          Hence, we have:
          $$
          begin{cases}
          begin{align}
          f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
          3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
          0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
          end{align}
          end{cases}
          $$
          Solve for $f(x)$:
          $$
          f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 14 '16 at 23:05

























          answered Oct 14 '16 at 22:59









          Kostiantyn Lapchevskyi

          1,221414




          1,221414























              0














              Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$



              Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$



              So we have three equations to work with:
              begin{align}
              f(x)+ 3f(1-tfrac {1}{x})= 7x,\
              f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
              f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
              end{align}



              Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
              which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$



              Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$



              which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$



              Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
              $$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$



              for $x ne 0,1.$



              As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
              for $x=1.$



              (Also note that $f(x) = k$ for $x=0$)






              share|cite|improve this answer




























                0














                Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$



                Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$



                So we have three equations to work with:
                begin{align}
                f(x)+ 3f(1-tfrac {1}{x})= 7x,\
                f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
                f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
                end{align}



                Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
                which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$



                Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$



                which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$



                Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
                $$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$



                for $x ne 0,1.$



                As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
                for $x=1.$



                (Also note that $f(x) = k$ for $x=0$)






                share|cite|improve this answer


























                  0












                  0








                  0






                  Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$



                  Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$



                  So we have three equations to work with:
                  begin{align}
                  f(x)+ 3f(1-tfrac {1}{x})= 7x,\
                  f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
                  f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
                  end{align}



                  Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
                  which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$



                  Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$



                  which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$



                  Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
                  $$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$



                  for $x ne 0,1.$



                  As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
                  for $x=1.$



                  (Also note that $f(x) = k$ for $x=0$)






                  share|cite|improve this answer














                  Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$



                  Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$



                  So we have three equations to work with:
                  begin{align}
                  f(x)+ 3f(1-tfrac {1}{x})= 7x,\
                  f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
                  f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
                  end{align}



                  Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
                  which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$



                  Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$



                  which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$



                  Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
                  $$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$



                  for $x ne 0,1.$



                  As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
                  for $x=1.$



                  (Also note that $f(x) = k$ for $x=0$)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 '18 at 20:22

























                  answered Nov 28 '18 at 5:37









                  BobL

                  11




                  11






























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