Difference between the absolute maximum and minimum of $f(x)=2x^3–9x^2+12x+5$












0












$begingroup$



Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:




The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.



This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.



Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    English hint: maxima and minima are plural. The singulars are maximum and minimum.
    $endgroup$
    – Ross Millikan
    Dec 25 '18 at 16:24






  • 1




    $begingroup$
    I think nine is the correct answer since the interval is closed.
    $endgroup$
    – Larry
    Dec 25 '18 at 16:26
















0












$begingroup$



Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:




The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.



This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.



Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    English hint: maxima and minima are plural. The singulars are maximum and minimum.
    $endgroup$
    – Ross Millikan
    Dec 25 '18 at 16:24






  • 1




    $begingroup$
    I think nine is the correct answer since the interval is closed.
    $endgroup$
    – Larry
    Dec 25 '18 at 16:26














0












0








0





$begingroup$



Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:




The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.



This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.



Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?






calculus







share|cite|improve this question











$endgroup$





Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:




The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.



This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.



Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 16:30









Bernard

121k740116




121k740116










asked Dec 25 '18 at 16:17









CaptainQuestionCaptainQuestion

1337




1337








  • 2




    $begingroup$
    English hint: maxima and minima are plural. The singulars are maximum and minimum.
    $endgroup$
    – Ross Millikan
    Dec 25 '18 at 16:24






  • 1




    $begingroup$
    I think nine is the correct answer since the interval is closed.
    $endgroup$
    – Larry
    Dec 25 '18 at 16:26














  • 2




    $begingroup$
    English hint: maxima and minima are plural. The singulars are maximum and minimum.
    $endgroup$
    – Ross Millikan
    Dec 25 '18 at 16:24






  • 1




    $begingroup$
    I think nine is the correct answer since the interval is closed.
    $endgroup$
    – Larry
    Dec 25 '18 at 16:26








2




2




$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24




$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24




1




1




$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26




$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26










3 Answers
3






active

oldest

votes


















3












$begingroup$

Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.



If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This not true, sorry. It is $$14-5=9$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 25 '18 at 16:25












  • $begingroup$
    @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
    $endgroup$
    – Ross Millikan
    Dec 25 '18 at 16:48





















3












$begingroup$

Sometimes a figure is worth a thousand words:



enter image description here






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.



    So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
    $$begin{array}{r!{}cccc}
    x=&0&1&2&3 \
    hline
    f(x)=&5&10&9&10
    end{array}$$
    We see the global maximum is $M=10$ and the global minimum is $m=5$, so
    $$M-m=5.*$$






    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%2f3052240%2fdifference-between-the-absolute-maximum-and-minimum-of-fx-2x3-9x212x5%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









      3












      $begingroup$

      Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.



      If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This not true, sorry. It is $$14-5=9$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 25 '18 at 16:25












      • $begingroup$
        @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
        $endgroup$
        – Ross Millikan
        Dec 25 '18 at 16:48


















      3












      $begingroup$

      Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.



      If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This not true, sorry. It is $$14-5=9$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 25 '18 at 16:25












      • $begingroup$
        @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
        $endgroup$
        – Ross Millikan
        Dec 25 '18 at 16:48
















      3












      3








      3





      $begingroup$

      Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.



      If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.






      share|cite|improve this answer









      $endgroup$



      Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.



      If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 25 '18 at 16:23









      Ross MillikanRoss Millikan

      296k23198371




      296k23198371












      • $begingroup$
        This not true, sorry. It is $$14-5=9$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 25 '18 at 16:25












      • $begingroup$
        @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
        $endgroup$
        – Ross Millikan
        Dec 25 '18 at 16:48




















      • $begingroup$
        This not true, sorry. It is $$14-5=9$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 25 '18 at 16:25












      • $begingroup$
        @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
        $endgroup$
        – Ross Millikan
        Dec 25 '18 at 16:48


















      $begingroup$
      This not true, sorry. It is $$14-5=9$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 25 '18 at 16:25






      $begingroup$
      This not true, sorry. It is $$14-5=9$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 25 '18 at 16:25














      $begingroup$
      @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
      $endgroup$
      – Ross Millikan
      Dec 25 '18 at 16:48






      $begingroup$
      @Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
      $endgroup$
      – Ross Millikan
      Dec 25 '18 at 16:48













      3












      $begingroup$

      Sometimes a figure is worth a thousand words:



      enter image description here






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Sometimes a figure is worth a thousand words:



        enter image description here






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Sometimes a figure is worth a thousand words:



          enter image description here






          share|cite|improve this answer









          $endgroup$



          Sometimes a figure is worth a thousand words:



          enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 16:43









          David G. StorkDavid G. Stork

          11k41432




          11k41432























              2












              $begingroup$

              The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.



              So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
              $$begin{array}{r!{}cccc}
              x=&0&1&2&3 \
              hline
              f(x)=&5&10&9&10
              end{array}$$
              We see the global maximum is $M=10$ and the global minimum is $m=5$, so
              $$M-m=5.*$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.



                So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
                $$begin{array}{r!{}cccc}
                x=&0&1&2&3 \
                hline
                f(x)=&5&10&9&10
                end{array}$$
                We see the global maximum is $M=10$ and the global minimum is $m=5$, so
                $$M-m=5.*$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.



                  So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
                  $$begin{array}{r!{}cccc}
                  x=&0&1&2&3 \
                  hline
                  f(x)=&5&10&9&10
                  end{array}$$
                  We see the global maximum is $M=10$ and the global minimum is $m=5$, so
                  $$M-m=5.*$$






                  share|cite|improve this answer









                  $endgroup$



                  The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.



                  So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
                  $$begin{array}{r!{}cccc}
                  x=&0&1&2&3 \
                  hline
                  f(x)=&5&10&9&10
                  end{array}$$
                  We see the global maximum is $M=10$ and the global minimum is $m=5$, so
                  $$M-m=5.*$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 17:05









                  BernardBernard

                  121k740116




                  121k740116






























                      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%2f3052240%2fdifference-between-the-absolute-maximum-and-minimum-of-fx-2x3-9x212x5%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

                      Bressuire

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                      Read more

                      Gyllenstierna

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more