A continuous function from (0,1) to [0,1]












1












$begingroup$


So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.



f(x) is a piece-wise function defined as:
begin{cases}
0, &quadtext{if x < 1/10}\
5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
1, &quadtext{if x > 9/10 }\
end{cases}



Thank you for your help and/or hints in prior.






real-analysis functions







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$endgroup$

















    1












    $begingroup$


    So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.



    f(x) is a piece-wise function defined as:
    begin{cases}
    0, &quadtext{if x < 1/10}\
    5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
    1, &quadtext{if x > 9/10 }\
    end{cases}



    Thank you for your help and/or hints in prior.






    real-analysis functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.



      f(x) is a piece-wise function defined as:
      begin{cases}
      0, &quadtext{if x < 1/10}\
      5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
      1, &quadtext{if x > 9/10 }\
      end{cases}



      Thank you for your help and/or hints in prior.






      real-analysis functions







      share|cite|improve this question











      $endgroup$




      So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.



      f(x) is a piece-wise function defined as:
      begin{cases}
      0, &quadtext{if x < 1/10}\
      5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
      1, &quadtext{if x > 9/10 }\
      end{cases}



      Thank you for your help and/or hints in prior.






      real-analysis functions




      real-analysis functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 16:43







      Allorja

















      asked Jan 8 at 16:13









      AllorjaAllorja

      789




      789






















          3 Answers
          3






          active

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          votes


















          1












          $begingroup$

          Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
          $$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$




          Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.



            Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              With the edit that I just made, my function is continuous right?
              $endgroup$
              – Allorja
              Jan 8 at 16:44






            • 1




              $begingroup$
              Yes, that ought to do it.
              $endgroup$
              – Arthur
              Jan 8 at 17:06



















            1












            $begingroup$

            Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
              $$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$




              Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
                $$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$




                Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
                  $$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$




                  Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.






                  share|cite|improve this answer









                  $endgroup$



                  Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
                  $$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$




                  Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 16:18









                  Hagen von EitzenHagen von Eitzen

                  280k23272504




                  280k23272504























                      3












                      $begingroup$

                      Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.



                      Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        With the edit that I just made, my function is continuous right?
                        $endgroup$
                        – Allorja
                        Jan 8 at 16:44






                      • 1




                        $begingroup$
                        Yes, that ought to do it.
                        $endgroup$
                        – Arthur
                        Jan 8 at 17:06
















                      3












                      $begingroup$

                      Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.



                      Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        With the edit that I just made, my function is continuous right?
                        $endgroup$
                        – Allorja
                        Jan 8 at 16:44






                      • 1




                        $begingroup$
                        Yes, that ought to do it.
                        $endgroup$
                        – Arthur
                        Jan 8 at 17:06














                      3












                      3








                      3





                      $begingroup$

                      Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.



                      Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.






                      share|cite|improve this answer









                      $endgroup$



                      Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.



                      Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 8 at 16:16









                      ArthurArthur

                      116k7116198




                      116k7116198












                      • $begingroup$
                        With the edit that I just made, my function is continuous right?
                        $endgroup$
                        – Allorja
                        Jan 8 at 16:44






                      • 1




                        $begingroup$
                        Yes, that ought to do it.
                        $endgroup$
                        – Arthur
                        Jan 8 at 17:06


















                      • $begingroup$
                        With the edit that I just made, my function is continuous right?
                        $endgroup$
                        – Allorja
                        Jan 8 at 16:44






                      • 1




                        $begingroup$
                        Yes, that ought to do it.
                        $endgroup$
                        – Arthur
                        Jan 8 at 17:06
















                      $begingroup$
                      With the edit that I just made, my function is continuous right?
                      $endgroup$
                      – Allorja
                      Jan 8 at 16:44




                      $begingroup$
                      With the edit that I just made, my function is continuous right?
                      $endgroup$
                      – Allorja
                      Jan 8 at 16:44




                      1




                      1




                      $begingroup$
                      Yes, that ought to do it.
                      $endgroup$
                      – Arthur
                      Jan 8 at 17:06




                      $begingroup$
                      Yes, that ought to do it.
                      $endgroup$
                      – Arthur
                      Jan 8 at 17:06











                      1












                      $begingroup$

                      Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.






                          share|cite|improve this answer









                          $endgroup$



                          Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 8 at 16:25









                          DubsDubs

                          55826




                          55826






























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