Maximum value of function $f(x)=frac{x^4-x^2}{x^6+2x^3-1}$ when $x >1$












1












$begingroup$


What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .



My try



Image 1



Unable to solve further.






real-analysis functions optimization maxima-minima a.m.-g.m.-inequality







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$endgroup$












  • $begingroup$
    What are $a$ and $b$ on the second page?
    $endgroup$
    – Matti P.
    Jan 16 at 11:43










  • $begingroup$
    Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Jan 16 at 11:46










  • $begingroup$
    $f(x)$ would be maximum when $1/f(x)$ would be minimum.
    $endgroup$
    – Paras Khosla
    Jan 16 at 14:11
















1












$begingroup$


What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .



My try



Image 1



Unable to solve further.






real-analysis functions optimization maxima-minima a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are $a$ and $b$ on the second page?
    $endgroup$
    – Matti P.
    Jan 16 at 11:43










  • $begingroup$
    Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Jan 16 at 11:46










  • $begingroup$
    $f(x)$ would be maximum when $1/f(x)$ would be minimum.
    $endgroup$
    – Paras Khosla
    Jan 16 at 14:11














1












1








1





$begingroup$


What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .



My try



Image 1



Unable to solve further.






real-analysis functions optimization maxima-minima a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$




What is the maximum value of the $$f(x)=frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .



My try



Image 1



Unable to solve further.






real-analysis functions optimization maxima-minima a.m.-g.m.-inequality




real-analysis functions optimization maxima-minima a.m.-g.m.-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 15:48







cattt

















asked Jan 16 at 11:40









catttcattt

242




242












  • $begingroup$
    What are $a$ and $b$ on the second page?
    $endgroup$
    – Matti P.
    Jan 16 at 11:43










  • $begingroup$
    Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Jan 16 at 11:46










  • $begingroup$
    $f(x)$ would be maximum when $1/f(x)$ would be minimum.
    $endgroup$
    – Paras Khosla
    Jan 16 at 14:11


















  • $begingroup$
    What are $a$ and $b$ on the second page?
    $endgroup$
    – Matti P.
    Jan 16 at 11:43










  • $begingroup$
    Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Jan 16 at 11:46










  • $begingroup$
    $f(x)$ would be maximum when $1/f(x)$ would be minimum.
    $endgroup$
    – Paras Khosla
    Jan 16 at 14:11
















$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43




$begingroup$
What are $a$ and $b$ on the second page?
$endgroup$
– Matti P.
Jan 16 at 11:43












$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46




$begingroup$
Hint: the derivative can be written as $$ f'(x) = frac{-2x(x^2 +1)(x^2-x-1)(x^4 + x^3 - x^2 - x -1)}{(x^6 + 2x^3 -1)^2} $$ Source: wolframalpha.com/input/…
$endgroup$
– Matti P.
Jan 16 at 11:46












$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11




$begingroup$
$f(x)$ would be maximum when $1/f(x)$ would be minimum.
$endgroup$
– Paras Khosla
Jan 16 at 14:11










1 Answer
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$begingroup$

Let $x-frac{1}{x}=t$.



Thus, $t>0$ and by AM-GM we obtain:
$$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
$$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.






share|cite|improve this answer









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    $begingroup$

    Let $x-frac{1}{x}=t$.



    Thus, $t>0$ and by AM-GM we obtain:
    $$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
    $$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
    The equality occurs for $t=1$, which says that we got a maximal value.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let $x-frac{1}{x}=t$.



      Thus, $t>0$ and by AM-GM we obtain:
      $$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
      $$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
      The equality occurs for $t=1$, which says that we got a maximal value.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $x-frac{1}{x}=t$.



        Thus, $t>0$ and by AM-GM we obtain:
        $$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
        $$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
        The equality occurs for $t=1$, which says that we got a maximal value.






        share|cite|improve this answer









        $endgroup$



        Let $x-frac{1}{x}=t$.



        Thus, $t>0$ and by AM-GM we obtain:
        $$frac{x^4-x^2}{x^6+2x^3-1}=frac{x-frac{1}{x}}{x^2-frac{1}{x^3}+2}=frac{t}{t^3+3t+2}=$$
        $$=frac{1}{t^2+frac{2}{t}+3}leqfrac{1}{3sqrt[3]{t^2left(frac{1}{t}right)^2}+3}=frac{1}{6}.$$
        The equality occurs for $t=1$, which says that we got a maximal value.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 13:49









        Michael RozenbergMichael Rozenberg

        104k1892197




        104k1892197






























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