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Investigating whether a function is bounded


How to determine whether to find an upper bound or lower boundProving Set bounded vs unboundedIs 1/(n+2) bounded or unbounded?Determine whether or not the improper integral converges $int_0^1 frace^xx^frac13 dx$Proving a set is boundedProve that the set of all almost upper bounds is nonempty, and bounded below.Finding upper bound of a linear functionDetermine whether a sequence is bounded aboveDetermine whether function is bounded$ < 0$ bounded? How to know?






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5















$begingroup$


I have the following function:



$$y=fracx^2-3x^2+7$$



and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.



However, how would I find the lower bound?










share|cite|improve this question











$endgroup$














  • $begingroup$
    You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
    $endgroup$
    – almagest
    Sep 29 at 12:59

















5















$begingroup$


I have the following function:



$$y=fracx^2-3x^2+7$$



and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.



However, how would I find the lower bound?










share|cite|improve this question











$endgroup$














  • $begingroup$
    You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
    $endgroup$
    – almagest
    Sep 29 at 12:59













5













5









5





$begingroup$


I have the following function:



$$y=fracx^2-3x^2+7$$



and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.



However, how would I find the lower bound?










share|cite|improve this question











$endgroup$




I have the following function:



$$y=fracx^2-3x^2+7$$



and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.



However, how would I find the lower bound?







real-analysis calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 29 at 13:08









cmk

7,2105 gold badges10 silver badges31 bronze badges




7,2105 gold badges10 silver badges31 bronze badges










asked Sep 29 at 12:37







user709660user709660





















  • $begingroup$
    You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
    $endgroup$
    – almagest
    Sep 29 at 12:59
















  • $begingroup$
    You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
    $endgroup$
    – almagest
    Sep 29 at 12:59















$begingroup$
You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
$endgroup$
– almagest
Sep 29 at 12:59




$begingroup$
You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
$endgroup$
– almagest
Sep 29 at 12:59










3 Answers
3






active

oldest

votes


















3

















$begingroup$

The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.






share|cite|improve this answer












$endgroup$





















    3

















    $begingroup$

    We have



    $$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$






    share|cite|improve this answer










    $endgroup$





















      1

















      $begingroup$

      $f(x)=fracx^2-3x^2+7$



      $=fracx^2+7x^2+7-frac10x^2+7$



      $=1-frac10x^2+7≥1-frac107=-frac37$



      $lim_x to infty f(x)=1$



      So,



      $1≥y≥-frac37$



      So the best upper bound is $1$.






      share|cite|improve this answer










      $endgroup$













      • $begingroup$
        It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
        $endgroup$
        – N. F. Taussig
        Sep 30 at 8:52










      • $begingroup$
        You're right. Thanks.
        $endgroup$
        – MENZIES
        Sep 30 at 8:53












      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3

















      $begingroup$

      The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.






      share|cite|improve this answer












      $endgroup$


















        3

















        $begingroup$

        The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.






        share|cite|improve this answer












        $endgroup$
















          3















          3











          3







          $begingroup$

          The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.






          share|cite|improve this answer












          $endgroup$



          The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.







          share|cite|improve this answer















          share|cite|improve this answer




          share|cite|improve this answer








          edited Sep 29 at 13:00

























          answered Sep 29 at 12:45









          mvpqmvpq

          83411 bronze badges




          83411 bronze badges


























              3

















              $begingroup$

              We have



              $$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$






              share|cite|improve this answer










              $endgroup$


















                3

















                $begingroup$

                We have



                $$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$






                share|cite|improve this answer










                $endgroup$
















                  3















                  3











                  3







                  $begingroup$

                  We have



                  $$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$






                  share|cite|improve this answer










                  $endgroup$



                  We have



                  $$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$







                  share|cite|improve this answer













                  share|cite|improve this answer




                  share|cite|improve this answer










                  answered Sep 29 at 13:13









                  useruser

                  112k10 gold badges49 silver badges103 bronze badges




                  112k10 gold badges49 silver badges103 bronze badges
























                      1

















                      $begingroup$

                      $f(x)=fracx^2-3x^2+7$



                      $=fracx^2+7x^2+7-frac10x^2+7$



                      $=1-frac10x^2+7≥1-frac107=-frac37$



                      $lim_x to infty f(x)=1$



                      So,



                      $1≥y≥-frac37$



                      So the best upper bound is $1$.






                      share|cite|improve this answer










                      $endgroup$













                      • $begingroup$
                        It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
                        $endgroup$
                        – N. F. Taussig
                        Sep 30 at 8:52










                      • $begingroup$
                        You're right. Thanks.
                        $endgroup$
                        – MENZIES
                        Sep 30 at 8:53















                      1

















                      $begingroup$

                      $f(x)=fracx^2-3x^2+7$



                      $=fracx^2+7x^2+7-frac10x^2+7$



                      $=1-frac10x^2+7≥1-frac107=-frac37$



                      $lim_x to infty f(x)=1$



                      So,



                      $1≥y≥-frac37$



                      So the best upper bound is $1$.






                      share|cite|improve this answer










                      $endgroup$













                      • $begingroup$
                        It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
                        $endgroup$
                        – N. F. Taussig
                        Sep 30 at 8:52










                      • $begingroup$
                        You're right. Thanks.
                        $endgroup$
                        – MENZIES
                        Sep 30 at 8:53













                      1















                      1











                      1







                      $begingroup$

                      $f(x)=fracx^2-3x^2+7$



                      $=fracx^2+7x^2+7-frac10x^2+7$



                      $=1-frac10x^2+7≥1-frac107=-frac37$



                      $lim_x to infty f(x)=1$



                      So,



                      $1≥y≥-frac37$



                      So the best upper bound is $1$.






                      share|cite|improve this answer










                      $endgroup$



                      $f(x)=fracx^2-3x^2+7$



                      $=fracx^2+7x^2+7-frac10x^2+7$



                      $=1-frac10x^2+7≥1-frac107=-frac37$



                      $lim_x to infty f(x)=1$



                      So,



                      $1≥y≥-frac37$



                      So the best upper bound is $1$.







                      share|cite|improve this answer













                      share|cite|improve this answer




                      share|cite|improve this answer










                      answered Sep 29 at 13:26









                      MENZIESMENZIES

                      437 bronze badges




                      437 bronze badges














                      • $begingroup$
                        It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
                        $endgroup$
                        – N. F. Taussig
                        Sep 30 at 8:52










                      • $begingroup$
                        You're right. Thanks.
                        $endgroup$
                        – MENZIES
                        Sep 30 at 8:53
















                      • $begingroup$
                        It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
                        $endgroup$
                        – N. F. Taussig
                        Sep 30 at 8:52










                      • $begingroup$
                        You're right. Thanks.
                        $endgroup$
                        – MENZIES
                        Sep 30 at 8:53















                      $begingroup$
                      It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
                      $endgroup$
                      – N. F. Taussig
                      Sep 30 at 8:52




                      $begingroup$
                      It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
                      $endgroup$
                      – N. F. Taussig
                      Sep 30 at 8:52












                      $begingroup$
                      You're right. Thanks.
                      $endgroup$
                      – MENZIES
                      Sep 30 at 8:53




                      $begingroup$
                      You're right. Thanks.
                      $endgroup$
                      – MENZIES
                      Sep 30 at 8:53


















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