How can I prove $frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$ using Rolle's theorem?











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For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.






analysis rolles-theorem







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up vote
0
down vote

favorite












For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.






analysis rolles-theorem







share|cite|improve this question









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onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • What have you already tried? Try editing your question to show this in order to get the most appropriate answers
    – MRobinson
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.






analysis rolles-theorem







share|cite|improve this question









New contributor




onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.






analysis rolles-theorem




analysis rolles-theorem






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onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • What have you already tried? Try editing your question to show this in order to get the most appropriate answers
    – MRobinson
    yesterday


















  • What have you already tried? Try editing your question to show this in order to get the most appropriate answers
    – MRobinson
    yesterday
















What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday




What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday










2 Answers
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Hint: I think you mean Cauchy mean value theorem
$$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
let $f(t)=t-arctan t$ and $g(t)=t^3$.






share|cite|improve this answer




























    up vote
    0
    down vote













    Hint:



    If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.






    share|cite|improve this answer





















    • This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
      – onzepattes
      yesterday











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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    active

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    up vote
    1
    down vote













    Hint: I think you mean Cauchy mean value theorem
    $$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
    let $f(t)=t-arctan t$ and $g(t)=t^3$.






    share|cite|improve this answer

























      up vote
      1
      down vote













      Hint: I think you mean Cauchy mean value theorem
      $$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
      let $f(t)=t-arctan t$ and $g(t)=t^3$.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Hint: I think you mean Cauchy mean value theorem
        $$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
        let $f(t)=t-arctan t$ and $g(t)=t^3$.






        share|cite|improve this answer












        Hint: I think you mean Cauchy mean value theorem
        $$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
        let $f(t)=t-arctan t$ and $g(t)=t^3$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Nosrati

        25.8k62252




        25.8k62252






















            up vote
            0
            down vote













            Hint:



            If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.






            share|cite|improve this answer





















            • This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
              – onzepattes
              yesterday















            up vote
            0
            down vote













            Hint:



            If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.






            share|cite|improve this answer





















            • This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
              – onzepattes
              yesterday













            up vote
            0
            down vote










            up vote
            0
            down vote









            Hint:



            If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.






            share|cite|improve this answer












            Hint:



            If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            ajotatxe

            52.1k23688




            52.1k23688












            • This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
              – onzepattes
              yesterday


















            • This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
              – onzepattes
              yesterday
















            This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
            – onzepattes
            yesterday




            This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
            – onzepattes
            yesterday










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