Upper Bound of $x(1-x)$ for $xin [0,1]$?











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Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?










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    The maximum value is $frac 1 4$
    – Kavi Rama Murthy
    Nov 14 at 8:05






  • 2




    For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
    – Yadati Kiran
    Nov 14 at 8:05






  • 3




    By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
    – Robert Z
    Nov 14 at 8:06












  • You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 15 at 13:54















up vote
2
down vote

favorite












Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?










share|cite|improve this question




















  • 2




    The maximum value is $frac 1 4$
    – Kavi Rama Murthy
    Nov 14 at 8:05






  • 2




    For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
    – Yadati Kiran
    Nov 14 at 8:05






  • 3




    By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
    – Robert Z
    Nov 14 at 8:06












  • You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 15 at 13:54













up vote
2
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up vote
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favorite











Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?










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Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?







real-analysis






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edited Nov 14 at 8:05

























asked Nov 14 at 8:03









Mog

2007




2007








  • 2




    The maximum value is $frac 1 4$
    – Kavi Rama Murthy
    Nov 14 at 8:05






  • 2




    For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
    – Yadati Kiran
    Nov 14 at 8:05






  • 3




    By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
    – Robert Z
    Nov 14 at 8:06












  • You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 15 at 13:54














  • 2




    The maximum value is $frac 1 4$
    – Kavi Rama Murthy
    Nov 14 at 8:05






  • 2




    For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
    – Yadati Kiran
    Nov 14 at 8:05






  • 3




    By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
    – Robert Z
    Nov 14 at 8:06












  • You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 15 at 13:54








2




2




The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05




The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05




2




2




For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05




For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05




3




3




By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06






By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06














You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54




You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54










4 Answers
4






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up vote
5
down vote



accepted










Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.



You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as



$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$






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    An upper bound is given by $frac{1}{4}$.



    Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.






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      up vote
      3
      down vote













      We have $x(1-x) ge 0$ and by AM-GM



      $$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$



      with equality for




      • $x=1-x implies x=frac12 $


      therefore as $x in[0,1]$



      $$0le x(1-x)le frac14$$






      share|cite|improve this answer




























        up vote
        2
        down vote













        Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

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          active

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          up vote
          5
          down vote



          accepted










          Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.



          You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as



          $$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$






          share|cite|improve this answer

























            up vote
            5
            down vote



            accepted










            Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.



            You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as



            $$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$






            share|cite|improve this answer























              up vote
              5
              down vote



              accepted







              up vote
              5
              down vote



              accepted






              Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.



              You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as



              $$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$






              share|cite|improve this answer












              Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.



              You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as



              $$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 14 at 8:06









              5xum

              88.3k392160




              88.3k392160






















                  up vote
                  5
                  down vote













                  An upper bound is given by $frac{1}{4}$.



                  Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.






                  share|cite|improve this answer








                  New contributor




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






















                    up vote
                    5
                    down vote













                    An upper bound is given by $frac{1}{4}$.



                    Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.






                    share|cite|improve this answer








                    New contributor




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




















                      up vote
                      5
                      down vote










                      up vote
                      5
                      down vote









                      An upper bound is given by $frac{1}{4}$.



                      Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.






                      share|cite|improve this answer








                      New contributor




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









                      An upper bound is given by $frac{1}{4}$.



                      Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.







                      share|cite|improve this answer








                      New contributor




                      Ekesh 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 answer



                      share|cite|improve this answer






                      New contributor




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









                      answered Nov 14 at 8:06









                      Ekesh

                      3414




                      3414




                      New contributor




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





                      New contributor





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






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






















                          up vote
                          3
                          down vote













                          We have $x(1-x) ge 0$ and by AM-GM



                          $$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$



                          with equality for




                          • $x=1-x implies x=frac12 $


                          therefore as $x in[0,1]$



                          $$0le x(1-x)le frac14$$






                          share|cite|improve this answer

























                            up vote
                            3
                            down vote













                            We have $x(1-x) ge 0$ and by AM-GM



                            $$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$



                            with equality for




                            • $x=1-x implies x=frac12 $


                            therefore as $x in[0,1]$



                            $$0le x(1-x)le frac14$$






                            share|cite|improve this answer























                              up vote
                              3
                              down vote










                              up vote
                              3
                              down vote









                              We have $x(1-x) ge 0$ and by AM-GM



                              $$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$



                              with equality for




                              • $x=1-x implies x=frac12 $


                              therefore as $x in[0,1]$



                              $$0le x(1-x)le frac14$$






                              share|cite|improve this answer












                              We have $x(1-x) ge 0$ and by AM-GM



                              $$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$



                              with equality for




                              • $x=1-x implies x=frac12 $


                              therefore as $x in[0,1]$



                              $$0le x(1-x)le frac14$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 14 at 8:18









                              gimusi

                              85.5k74294




                              85.5k74294






















                                  up vote
                                  2
                                  down vote













                                  Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.






                                  share|cite|improve this answer

























                                    up vote
                                    2
                                    down vote













                                    Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.






                                    share|cite|improve this answer























                                      up vote
                                      2
                                      down vote










                                      up vote
                                      2
                                      down vote









                                      Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.






                                      share|cite|improve this answer












                                      Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 14 at 8:12









                                      Yadati Kiran

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