Finding the minimum value of a function without using Calculus












8












$begingroup$



Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$




My attempt:



Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.



I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)



Any help would be appreciated.






functions maxima-minima







share|cite|improve this question











$endgroup$

















    8












    $begingroup$



    Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$




    My attempt:



    Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.



    I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)



    Any help would be appreciated.






    functions maxima-minima







    share|cite|improve this question











    $endgroup$















      8












      8








      8


      2



      $begingroup$



      Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$




      My attempt:



      Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.



      I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)



      Any help would be appreciated.






      functions maxima-minima







      share|cite|improve this question











      $endgroup$





      Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$




      My attempt:



      Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.



      I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)



      Any help would be appreciated.






      functions maxima-minima




      functions maxima-minima






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 9 at 14:06









      Bernard

      123k741116




      123k741116










      asked Mar 9 at 13:53









      TonyTony

      907




      907






















          3 Answers
          3






          active

          oldest

          votes


















          13












          $begingroup$

          Use AM-GM:
          $$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
          equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.






          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            Hint:



            As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality



            $$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$



            Set $4a-2b=0iff b=2a$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
              $endgroup$
              – lab bhattacharjee
              Mar 9 at 17:48



















            2












            $begingroup$

            First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
            $$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
            has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.



            The discriminant of $(x^2)^3 - b(x^2) + 1$ is
            $$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
            The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3141146%2ffinding-the-minimum-value-of-a-function-without-using-calculus%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              13












              $begingroup$

              Use AM-GM:
              $$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
              equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.






              share|cite|improve this answer









              $endgroup$


















                13












                $begingroup$

                Use AM-GM:
                $$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
                equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.






                share|cite|improve this answer









                $endgroup$
















                  13












                  13








                  13





                  $begingroup$

                  Use AM-GM:
                  $$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
                  equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.






                  share|cite|improve this answer









                  $endgroup$



                  Use AM-GM:
                  $$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
                  equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 9 at 14:00









                  farruhotafarruhota

                  21.1k2841




                  21.1k2841























                      5












                      $begingroup$

                      Hint:



                      As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality



                      $$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$



                      Set $4a-2b=0iff b=2a$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
                        $endgroup$
                        – lab bhattacharjee
                        Mar 9 at 17:48
















                      5












                      $begingroup$

                      Hint:



                      As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality



                      $$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$



                      Set $4a-2b=0iff b=2a$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
                        $endgroup$
                        – lab bhattacharjee
                        Mar 9 at 17:48














                      5












                      5








                      5





                      $begingroup$

                      Hint:



                      As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality



                      $$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$



                      Set $4a-2b=0iff b=2a$






                      share|cite|improve this answer











                      $endgroup$



                      Hint:



                      As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality



                      $$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$



                      Set $4a-2b=0iff b=2a$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Mar 9 at 15:56

























                      answered Mar 9 at 13:58









                      lab bhattacharjeelab bhattacharjee

                      227k15158275




                      227k15158275












                      • $begingroup$
                        @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
                        $endgroup$
                        – lab bhattacharjee
                        Mar 9 at 17:48


















                      • $begingroup$
                        @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
                        $endgroup$
                        – lab bhattacharjee
                        Mar 9 at 17:48
















                      $begingroup$
                      @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
                      $endgroup$
                      – lab bhattacharjee
                      Mar 9 at 17:48




                      $begingroup$
                      @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
                      $endgroup$
                      – lab bhattacharjee
                      Mar 9 at 17:48











                      2












                      $begingroup$

                      First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
                      $$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
                      has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.



                      The discriminant of $(x^2)^3 - b(x^2) + 1$ is
                      $$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
                      The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
                        $$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
                        has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.



                        The discriminant of $(x^2)^3 - b(x^2) + 1$ is
                        $$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
                        The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
                          $$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
                          has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.



                          The discriminant of $(x^2)^3 - b(x^2) + 1$ is
                          $$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
                          The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.






                          share|cite|improve this answer









                          $endgroup$



                          First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
                          $$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
                          has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.



                          The discriminant of $(x^2)^3 - b(x^2) + 1$ is
                          $$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
                          The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 10 at 0:22









                          Eric TowersEric Towers

                          32.9k22370




                          32.9k22370






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3141146%2ffinding-the-minimum-value-of-a-function-without-using-calculus%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              How to change which sound is reproduced for terminal bell?

                              Image
                              3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                              Read more

                              Can I use Tabulator js library in my java Spring + Thymeleaf project?

                              Image
                              0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                              Read more

                              Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                              Image
                              up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                              Read more