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Is it appropriate to consider a hole in the graph a zero?


Can the graph of $x^x$ have a real-valued plot below zero?How to evaluate a zero of the Riemann zeta function?Graph the function and apply the appropriate transformations $y = (1)/( x + 4)$Converting a graph into a functionWhy don't graphing tools represent holes in a graph?Calculating equation for a graphGraph of a Function or a Just a Relation?Take the function $f(x)=8x+120$. Find a function $g(x)$ which contains the point $(0,0)$ and is asymptotic to $f(x)$.Consider the family of equations $y=x^n$ with the integer parameter n.When is it appropriate to neglect the arguments of the function?













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$begingroup$


Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
    It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
      It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?










      share|cite|improve this question









      $endgroup$




      Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
      It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?







      functions terminology roots graphing-functions






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      AnadactotheAnadactothe

      13211




      13211






















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          1












          $begingroup$

          A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.



          On the other hand, since the discontinuity is removable, we can plug the hole by defining
          $$g(x)=begin{cases}
          f(x), &x neq 0\
          0, &x=0
          end{cases}$$

          in which case $x=0$ is indeed a zero of $g(x)$.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.



            On the other hand, since the discontinuity is removable, we can plug the hole by defining
            $$g(x)=begin{cases}
            f(x), &x neq 0\
            0, &x=0
            end{cases}$$

            in which case $x=0$ is indeed a zero of $g(x)$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.



              On the other hand, since the discontinuity is removable, we can plug the hole by defining
              $$g(x)=begin{cases}
              f(x), &x neq 0\
              0, &x=0
              end{cases}$$

              in which case $x=0$ is indeed a zero of $g(x)$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.



                On the other hand, since the discontinuity is removable, we can plug the hole by defining
                $$g(x)=begin{cases}
                f(x), &x neq 0\
                0, &x=0
                end{cases}$$

                in which case $x=0$ is indeed a zero of $g(x)$.






                share|cite|improve this answer









                $endgroup$



                A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.



                On the other hand, since the discontinuity is removable, we can plug the hole by defining
                $$g(x)=begin{cases}
                f(x), &x neq 0\
                0, &x=0
                end{cases}$$

                in which case $x=0$ is indeed a zero of $g(x)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                ThéophileThéophile

                20.2k13047




                20.2k13047






























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