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
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          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









            $endgroup$













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              3 Answers
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              active

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              3 Answers
              3






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              active

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              active

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              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






























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