Showing that $f(x)$ is convex in $(0,3)$












1














I've got the following function: $$f(x)=frac{1}{16x}-frac{1}{(x+3)^2} $$
And I wish to show that it is convex in the open interval $(0,3)$, took the second derivative, i.e.$$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}$$
I painstakingly tried to expand it and algebraically manipulate it to solve $f''(x)>0$, but that seems pretty hopeless, how should I go about this?






calculus inequality functional-inequalities







share|cite|improve this question



























    1














    I've got the following function: $$f(x)=frac{1}{16x}-frac{1}{(x+3)^2} $$
    And I wish to show that it is convex in the open interval $(0,3)$, took the second derivative, i.e.$$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}$$
    I painstakingly tried to expand it and algebraically manipulate it to solve $f''(x)>0$, but that seems pretty hopeless, how should I go about this?






    calculus inequality functional-inequalities







    share|cite|improve this question

























      1












      1








      1







      I've got the following function: $$f(x)=frac{1}{16x}-frac{1}{(x+3)^2} $$
      And I wish to show that it is convex in the open interval $(0,3)$, took the second derivative, i.e.$$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}$$
      I painstakingly tried to expand it and algebraically manipulate it to solve $f''(x)>0$, but that seems pretty hopeless, how should I go about this?






      calculus inequality functional-inequalities







      share|cite|improve this question













      I've got the following function: $$f(x)=frac{1}{16x}-frac{1}{(x+3)^2} $$
      And I wish to show that it is convex in the open interval $(0,3)$, took the second derivative, i.e.$$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}$$
      I painstakingly tried to expand it and algebraically manipulate it to solve $f''(x)>0$, but that seems pretty hopeless, how should I go about this?






      calculus inequality functional-inequalities




      calculus inequality functional-inequalities






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 24 at 18:07









      Spasoje Durovic

      19410




      19410






















          2 Answers
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          2














          We are claiming that



          $$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}>0 ifffrac{(x+3)^4-48x^3}{8x^3(x+3)^4}>0$$



          then we need to prove that (using the idea by Timothy for the rational roots)



          $$bar g(x)=(x+3)^4-48x^3=(x-3)(x^3-33x^2-45x-27)>0$$



          that is



          $$g(x)=x^3-33x^2-45x-27<0$$



          and we have




          • $g(0)=-27$

          • $g(3)=-432$

          • $h(x)=g'(x)=3x^2-66x-45<0$


          therefore





          • $g(x)$ is strictly decreasing on the interval


          • $g(x)$ is negative on the interval






          share|cite|improve this answer































            1














            Here's something I would do:



            Solve $f''(x)=0$ and test a value inside your open interval.



            So we have: $$frac1{8x^3}-frac6{(x+3)^4}=0$$



            $$frac{1}{8x^3}=frac{6}{(x+3)^4}$$



            Then I would cross multiply and expand $(x+3)^4$:



            $$(x+3)^4=48x^3$$



            $$x^4+12x^3+54x^2+108x+81=48x^3$$
            $$x^4-36x^3+54x^2+108x+81=0$$



            At this point, we could use the Rational Root Theorem to guess roots.



            The possible roots are $pm1, pm3, pm27, pm81$.



            Of these, we find that $x=3$ works, as $3^4-36(3)^3+54(3)^2+108(3)+81=0$.



            I think you could take it from here.






            share|cite|improve this answer





















            • Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
              – Spasoje Durovic
              Nov 24 at 18:33










            • yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
              – Timothy Cho
              Nov 24 at 18:33








            • 1




              RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
              – gimusi
              Nov 24 at 18:40












            • I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
              – Spasoje Durovic
              Nov 24 at 18:46






            • 1




              @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
              – gimusi
              Nov 24 at 19:16











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            2 Answers
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            2 Answers
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            2














            We are claiming that



            $$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}>0 ifffrac{(x+3)^4-48x^3}{8x^3(x+3)^4}>0$$



            then we need to prove that (using the idea by Timothy for the rational roots)



            $$bar g(x)=(x+3)^4-48x^3=(x-3)(x^3-33x^2-45x-27)>0$$



            that is



            $$g(x)=x^3-33x^2-45x-27<0$$



            and we have




            • $g(0)=-27$

            • $g(3)=-432$

            • $h(x)=g'(x)=3x^2-66x-45<0$


            therefore





            • $g(x)$ is strictly decreasing on the interval


            • $g(x)$ is negative on the interval






            share|cite|improve this answer




























              2














              We are claiming that



              $$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}>0 ifffrac{(x+3)^4-48x^3}{8x^3(x+3)^4}>0$$



              then we need to prove that (using the idea by Timothy for the rational roots)



              $$bar g(x)=(x+3)^4-48x^3=(x-3)(x^3-33x^2-45x-27)>0$$



              that is



              $$g(x)=x^3-33x^2-45x-27<0$$



              and we have




              • $g(0)=-27$

              • $g(3)=-432$

              • $h(x)=g'(x)=3x^2-66x-45<0$


              therefore





              • $g(x)$ is strictly decreasing on the interval


              • $g(x)$ is negative on the interval






              share|cite|improve this answer


























                2












                2








                2






                We are claiming that



                $$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}>0 ifffrac{(x+3)^4-48x^3}{8x^3(x+3)^4}>0$$



                then we need to prove that (using the idea by Timothy for the rational roots)



                $$bar g(x)=(x+3)^4-48x^3=(x-3)(x^3-33x^2-45x-27)>0$$



                that is



                $$g(x)=x^3-33x^2-45x-27<0$$



                and we have




                • $g(0)=-27$

                • $g(3)=-432$

                • $h(x)=g'(x)=3x^2-66x-45<0$


                therefore





                • $g(x)$ is strictly decreasing on the interval


                • $g(x)$ is negative on the interval






                share|cite|improve this answer














                We are claiming that



                $$f''(x)=frac{1}{8x^3}-frac{6}{(x+3)^4}>0 ifffrac{(x+3)^4-48x^3}{8x^3(x+3)^4}>0$$



                then we need to prove that (using the idea by Timothy for the rational roots)



                $$bar g(x)=(x+3)^4-48x^3=(x-3)(x^3-33x^2-45x-27)>0$$



                that is



                $$g(x)=x^3-33x^2-45x-27<0$$



                and we have




                • $g(0)=-27$

                • $g(3)=-432$

                • $h(x)=g'(x)=3x^2-66x-45<0$


                therefore





                • $g(x)$ is strictly decreasing on the interval


                • $g(x)$ is negative on the interval







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 24 at 19:06

























                answered Nov 24 at 18:34









                gimusi

                1




                1























                    1














                    Here's something I would do:



                    Solve $f''(x)=0$ and test a value inside your open interval.



                    So we have: $$frac1{8x^3}-frac6{(x+3)^4}=0$$



                    $$frac{1}{8x^3}=frac{6}{(x+3)^4}$$



                    Then I would cross multiply and expand $(x+3)^4$:



                    $$(x+3)^4=48x^3$$



                    $$x^4+12x^3+54x^2+108x+81=48x^3$$
                    $$x^4-36x^3+54x^2+108x+81=0$$



                    At this point, we could use the Rational Root Theorem to guess roots.



                    The possible roots are $pm1, pm3, pm27, pm81$.



                    Of these, we find that $x=3$ works, as $3^4-36(3)^3+54(3)^2+108(3)+81=0$.



                    I think you could take it from here.






                    share|cite|improve this answer





















                    • Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
                      – Spasoje Durovic
                      Nov 24 at 18:33










                    • yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
                      – Timothy Cho
                      Nov 24 at 18:33








                    • 1




                      RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
                      – gimusi
                      Nov 24 at 18:40












                    • I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
                      – Spasoje Durovic
                      Nov 24 at 18:46






                    • 1




                      @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
                      – gimusi
                      Nov 24 at 19:16
















                    1














                    Here's something I would do:



                    Solve $f''(x)=0$ and test a value inside your open interval.



                    So we have: $$frac1{8x^3}-frac6{(x+3)^4}=0$$



                    $$frac{1}{8x^3}=frac{6}{(x+3)^4}$$



                    Then I would cross multiply and expand $(x+3)^4$:



                    $$(x+3)^4=48x^3$$



                    $$x^4+12x^3+54x^2+108x+81=48x^3$$
                    $$x^4-36x^3+54x^2+108x+81=0$$



                    At this point, we could use the Rational Root Theorem to guess roots.



                    The possible roots are $pm1, pm3, pm27, pm81$.



                    Of these, we find that $x=3$ works, as $3^4-36(3)^3+54(3)^2+108(3)+81=0$.



                    I think you could take it from here.






                    share|cite|improve this answer





















                    • Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
                      – Spasoje Durovic
                      Nov 24 at 18:33










                    • yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
                      – Timothy Cho
                      Nov 24 at 18:33








                    • 1




                      RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
                      – gimusi
                      Nov 24 at 18:40












                    • I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
                      – Spasoje Durovic
                      Nov 24 at 18:46






                    • 1




                      @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
                      – gimusi
                      Nov 24 at 19:16














                    1












                    1








                    1






                    Here's something I would do:



                    Solve $f''(x)=0$ and test a value inside your open interval.



                    So we have: $$frac1{8x^3}-frac6{(x+3)^4}=0$$



                    $$frac{1}{8x^3}=frac{6}{(x+3)^4}$$



                    Then I would cross multiply and expand $(x+3)^4$:



                    $$(x+3)^4=48x^3$$



                    $$x^4+12x^3+54x^2+108x+81=48x^3$$
                    $$x^4-36x^3+54x^2+108x+81=0$$



                    At this point, we could use the Rational Root Theorem to guess roots.



                    The possible roots are $pm1, pm3, pm27, pm81$.



                    Of these, we find that $x=3$ works, as $3^4-36(3)^3+54(3)^2+108(3)+81=0$.



                    I think you could take it from here.






                    share|cite|improve this answer












                    Here's something I would do:



                    Solve $f''(x)=0$ and test a value inside your open interval.



                    So we have: $$frac1{8x^3}-frac6{(x+3)^4}=0$$



                    $$frac{1}{8x^3}=frac{6}{(x+3)^4}$$



                    Then I would cross multiply and expand $(x+3)^4$:



                    $$(x+3)^4=48x^3$$



                    $$x^4+12x^3+54x^2+108x+81=48x^3$$
                    $$x^4-36x^3+54x^2+108x+81=0$$



                    At this point, we could use the Rational Root Theorem to guess roots.



                    The possible roots are $pm1, pm3, pm27, pm81$.



                    Of these, we find that $x=3$ works, as $3^4-36(3)^3+54(3)^2+108(3)+81=0$.



                    I think you could take it from here.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 at 18:30









                    Timothy Cho

                    789519




                    789519












                    • Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
                      – Spasoje Durovic
                      Nov 24 at 18:33










                    • yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
                      – Timothy Cho
                      Nov 24 at 18:33








                    • 1




                      RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
                      – gimusi
                      Nov 24 at 18:40












                    • I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
                      – Spasoje Durovic
                      Nov 24 at 18:46






                    • 1




                      @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
                      – gimusi
                      Nov 24 at 19:16


















                    • Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
                      – Spasoje Durovic
                      Nov 24 at 18:33










                    • yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
                      – Timothy Cho
                      Nov 24 at 18:33








                    • 1




                      RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
                      – gimusi
                      Nov 24 at 18:40












                    • I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
                      – Spasoje Durovic
                      Nov 24 at 18:46






                    • 1




                      @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
                      – gimusi
                      Nov 24 at 19:16
















                    Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
                    – Spasoje Durovic
                    Nov 24 at 18:33




                    Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.-
                    – Spasoje Durovic
                    Nov 24 at 18:33












                    yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
                    – Timothy Cho
                    Nov 24 at 18:33






                    yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3
                    – Timothy Cho
                    Nov 24 at 18:33






                    1




                    1




                    RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
                    – gimusi
                    Nov 24 at 18:40






                    RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$?
                    – gimusi
                    Nov 24 at 18:40














                    I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
                    – Spasoje Durovic
                    Nov 24 at 18:46




                    I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it
                    – Spasoje Durovic
                    Nov 24 at 18:46




                    1




                    1




                    @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
                    – gimusi
                    Nov 24 at 19:16




                    @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval.
                    – gimusi
                    Nov 24 at 19:16


















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                    Puebla de Zaragoza

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                    Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                    Musa

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                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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