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












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$begingroup$



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







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  • 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






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edited Dec 25 '18 at 16:30









Bernard

121k740116




121k740116










asked Dec 25 '18 at 16:17









CaptainQuestionCaptainQuestion

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  • 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









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      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

      297k23198371




      297k23198371












      • $begingroup$
        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






























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