Why does simplifying a function change its domain?












1












$begingroup$


Perhaps this is a silly question, but if you have a function, such as



$$f(x) = frac{x^2}{x}$$



the domain is all real numbers except x = 0.



However, this function simplifies to



$$f(x) = x$$



which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?






functions algebras







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$endgroup$












  • $begingroup$
    To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
    $endgroup$
    – glglgl
    May 25 '14 at 5:54






  • 2




    $begingroup$
    Possible duplicate of Why does factoring eliminate a hole in the limit?
    $endgroup$
    – Alex M.
    Sep 14 '16 at 21:56
















1












$begingroup$


Perhaps this is a silly question, but if you have a function, such as



$$f(x) = frac{x^2}{x}$$



the domain is all real numbers except x = 0.



However, this function simplifies to



$$f(x) = x$$



which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?






functions algebras







share|cite|improve this question









$endgroup$












  • $begingroup$
    To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
    $endgroup$
    – glglgl
    May 25 '14 at 5:54






  • 2




    $begingroup$
    Possible duplicate of Why does factoring eliminate a hole in the limit?
    $endgroup$
    – Alex M.
    Sep 14 '16 at 21:56














1












1








1





$begingroup$


Perhaps this is a silly question, but if you have a function, such as



$$f(x) = frac{x^2}{x}$$



the domain is all real numbers except x = 0.



However, this function simplifies to



$$f(x) = x$$



which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?






functions algebras







share|cite|improve this question









$endgroup$




Perhaps this is a silly question, but if you have a function, such as



$$f(x) = frac{x^2}{x}$$



the domain is all real numbers except x = 0.



However, this function simplifies to



$$f(x) = x$$



which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?






functions algebras




functions algebras






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 25 '14 at 3:17









Jack HumphriesJack Humphries

19110




19110












  • $begingroup$
    To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
    $endgroup$
    – glglgl
    May 25 '14 at 5:54






  • 2




    $begingroup$
    Possible duplicate of Why does factoring eliminate a hole in the limit?
    $endgroup$
    – Alex M.
    Sep 14 '16 at 21:56


















  • $begingroup$
    To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
    $endgroup$
    – glglgl
    May 25 '14 at 5:54






  • 2




    $begingroup$
    Possible duplicate of Why does factoring eliminate a hole in the limit?
    $endgroup$
    – Alex M.
    Sep 14 '16 at 21:56
















$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54




$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54




2




2




$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56




$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56










2 Answers
2






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oldest

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$begingroup$

To be strictly correct, the domain does not change. The simplified version of
$$f(x)=frac{x^2}{x} ,quad xne0$$
is
$$f(x)=x ,quad xne0 .$$



Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.



    If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10












      $begingroup$

      To be strictly correct, the domain does not change. The simplified version of
      $$f(x)=frac{x^2}{x} ,quad xne0$$
      is
      $$f(x)=x ,quad xne0 .$$



      Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.






      share|cite|improve this answer









      $endgroup$


















        10












        $begingroup$

        To be strictly correct, the domain does not change. The simplified version of
        $$f(x)=frac{x^2}{x} ,quad xne0$$
        is
        $$f(x)=x ,quad xne0 .$$



        Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.






        share|cite|improve this answer









        $endgroup$
















          10












          10








          10





          $begingroup$

          To be strictly correct, the domain does not change. The simplified version of
          $$f(x)=frac{x^2}{x} ,quad xne0$$
          is
          $$f(x)=x ,quad xne0 .$$



          Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.






          share|cite|improve this answer









          $endgroup$



          To be strictly correct, the domain does not change. The simplified version of
          $$f(x)=frac{x^2}{x} ,quad xne0$$
          is
          $$f(x)=x ,quad xne0 .$$



          Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 25 '14 at 3:33









          DavidDavid

          68.9k667130




          68.9k667130























              -1












              $begingroup$

              Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.



              If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.



                If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.



                  If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$






                  share|cite|improve this answer











                  $endgroup$



                  Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.



                  If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited May 26 '14 at 22:48









                  user7000

                  2791822




                  2791822










                  answered May 25 '14 at 5:22









                  user121955user121955

                  1387




                  1387






























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