Maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$What's the minimum value?Finding the maximum and minimum values of $f(x)=a^x+a^1/x$notation for minimum and maximum?Piecewise function and maximum/minMinimum of a maximum functionMaximum value of function given minimum valueMaximum and minimum for piecewise functionMaximum and minimum values of functionStrictly convex minimum and maximumFinding the minimum value of a function without using Calculus

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Maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$


What's the minimum value?Finding the maximum and minimum values of $f(x)=a^x+a^1/x$notation for minimum and maximum?Piecewise function and maximum/minMinimum of a maximum functionMaximum value of function given minimum valueMaximum and minimum for piecewise functionMaximum and minimum values of functionStrictly convex minimum and maximumFinding the minimum value of a function without using Calculus













1












$begingroup$


Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.



I tried to solve this problem using the value of $y'=0$.



Solving the I got $y=fracax+bAx+B$.



Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.










share|cite|improve this question









$endgroup$











  • $begingroup$
    For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
    $endgroup$
    – Dave L. Renfro
    Mar 13 at 11:08















1












$begingroup$


Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.



I tried to solve this problem using the value of $y'=0$.



Solving the I got $y=fracax+bAx+B$.



Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.










share|cite|improve this question









$endgroup$











  • $begingroup$
    For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
    $endgroup$
    – Dave L. Renfro
    Mar 13 at 11:08













1












1








1





$begingroup$


Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.



I tried to solve this problem using the value of $y'=0$.



Solving the I got $y=fracax+bAx+B$.



Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.










share|cite|improve this question









$endgroup$




Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.



I tried to solve this problem using the value of $y'=0$.



Solving the I got $y=fracax+bAx+B$.



Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.







functions






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asked Mar 13 at 10:44









Samar Imam ZaidiSamar Imam Zaidi

1,5751520




1,5751520











  • $begingroup$
    For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
    $endgroup$
    – Dave L. Renfro
    Mar 13 at 11:08
















  • $begingroup$
    For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
    $endgroup$
    – Dave L. Renfro
    Mar 13 at 11:08















$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08




$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08










1 Answer
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0












$begingroup$

Hint: By the quotient rule we get
$$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
C right) ^2
$$






share|cite|improve this answer









$endgroup$












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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint: By the quotient rule we get
    $$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
    C right) ^2
    $$






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      Hint: By the quotient rule we get
      $$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
      C right) ^2
      $$






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Hint: By the quotient rule we get
        $$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
        C right) ^2
        $$






        share|cite|improve this answer









        $endgroup$



        Hint: By the quotient rule we get
        $$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
        C right) ^2
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 13 at 11:45









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        77.9k42866




        77.9k42866



























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