Maximum Value of $g(x)=(8+x)^3(8-x)^4$












1














I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?






functions







share|cite|improve this question
























  • Yes Moo. Thank you very much.
    – Sherlock Watson
    Nov 28 '18 at 5:40










  • There is a local maximum and no global maximum - do a plot to see that.
    – Moo
    Nov 28 '18 at 5:41










  • I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
    – Sherlock Watson
    Nov 28 '18 at 5:44










  • That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
    – Eevee Trainer
    Nov 28 '18 at 5:44






  • 1




    Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:49
















1














I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?






functions







share|cite|improve this question
























  • Yes Moo. Thank you very much.
    – Sherlock Watson
    Nov 28 '18 at 5:40










  • There is a local maximum and no global maximum - do a plot to see that.
    – Moo
    Nov 28 '18 at 5:41










  • I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
    – Sherlock Watson
    Nov 28 '18 at 5:44










  • That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
    – Eevee Trainer
    Nov 28 '18 at 5:44






  • 1




    Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:49














1












1








1


0





I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?






functions







share|cite|improve this question















I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 5:39









Moo

5,53131020




5,53131020










asked Nov 28 '18 at 5:38









Sherlock WatsonSherlock Watson

3422413




3422413












  • Yes Moo. Thank you very much.
    – Sherlock Watson
    Nov 28 '18 at 5:40










  • There is a local maximum and no global maximum - do a plot to see that.
    – Moo
    Nov 28 '18 at 5:41










  • I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
    – Sherlock Watson
    Nov 28 '18 at 5:44










  • That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
    – Eevee Trainer
    Nov 28 '18 at 5:44






  • 1




    Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:49


















  • Yes Moo. Thank you very much.
    – Sherlock Watson
    Nov 28 '18 at 5:40










  • There is a local maximum and no global maximum - do a plot to see that.
    – Moo
    Nov 28 '18 at 5:41










  • I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
    – Sherlock Watson
    Nov 28 '18 at 5:44










  • That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
    – Eevee Trainer
    Nov 28 '18 at 5:44






  • 1




    Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:49
















Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40




Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40












There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41




There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41












I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44




I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44












That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44




That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44




1




1




Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49




Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49










3 Answers
3






active

oldest

votes


















1














$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,



with leading term $x^7$.



1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$



2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.



3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).






share|cite|improve this answer































    0














    We have that



    $$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$



    $$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$



    which is true when




    • $x=pm 8$

    • $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$


    and the latter is a local maximum.






    share|cite|improve this answer





























      0














      You can use the wavy curve approach to find the range in which the local extremas and global extremas.



      The wavy curve method is specifically used for polynomial equations.



      Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7






      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









        1














        $p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,



        with leading term $x^7$.



        1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$



        2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.



        3)$ -8 <x < 8$ : $p(x) >0$.
        (Has a relative maximum).






        share|cite|improve this answer




























          1














          $p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,



          with leading term $x^7$.



          1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$



          2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.



          3)$ -8 <x < 8$ : $p(x) >0$.
          (Has a relative maximum).






          share|cite|improve this answer


























            1












            1








            1






            $p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,



            with leading term $x^7$.



            1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$



            2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.



            3)$ -8 <x < 8$ : $p(x) >0$.
            (Has a relative maximum).






            share|cite|improve this answer














            $p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,



            with leading term $x^7$.



            1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$



            2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.



            3)$ -8 <x < 8$ : $p(x) >0$.
            (Has a relative maximum).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 '18 at 7:44

























            answered Nov 28 '18 at 7:28









            Peter SzilasPeter Szilas

            10.9k2720




            10.9k2720























                0














                We have that



                $$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$



                $$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$



                which is true when




                • $x=pm 8$

                • $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$


                and the latter is a local maximum.






                share|cite|improve this answer


























                  0














                  We have that



                  $$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$



                  $$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$



                  which is true when




                  • $x=pm 8$

                  • $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$


                  and the latter is a local maximum.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    We have that



                    $$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$



                    $$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$



                    which is true when




                    • $x=pm 8$

                    • $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$


                    and the latter is a local maximum.






                    share|cite|improve this answer












                    We have that



                    $$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$



                    $$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$



                    which is true when




                    • $x=pm 8$

                    • $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$


                    and the latter is a local maximum.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 '18 at 11:23









                    gimusigimusi

                    1




                    1























                        0














                        You can use the wavy curve approach to find the range in which the local extremas and global extremas.



                        The wavy curve method is specifically used for polynomial equations.



                        Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7






                        share|cite|improve this answer


























                          0














                          You can use the wavy curve approach to find the range in which the local extremas and global extremas.



                          The wavy curve method is specifically used for polynomial equations.



                          Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7






                          share|cite|improve this answer
























                            0












                            0








                            0






                            You can use the wavy curve approach to find the range in which the local extremas and global extremas.



                            The wavy curve method is specifically used for polynomial equations.



                            Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7






                            share|cite|improve this answer












                            You can use the wavy curve approach to find the range in which the local extremas and global extremas.



                            The wavy curve method is specifically used for polynomial equations.



                            Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 28 '18 at 11:43









                            Bhargav KaleBhargav Kale

                            313




                            313






























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