If $f(x)=int_{0}^{x}sqrt {f(t)}dt$, then find $f(6)$.












3












$begingroup$


Let $f:[0,infty)to[0,infty)$ be continuous on $[0,infty)$ and differentiable on $(0,infty)$. If $f(x)=int_{0}^{x}sqrt {f(t)}dt$, then find $f(6)$.
$$f(x)=int_{0}^{x}sqrt {f(t)}dt$$
$$g(x):=sqrt {f(x)}implies(g(x))^2=f(x)=int_{0}^{x}g(t)dtimplies2g(x)g'(x)=g(x)$$
[by FTC-1]
$$implies g=0 vee g'(x)=frac{1}{2} $$
$$(g(0))^2=f(0)=0implies g(x)=frac{x}{2}vee g=0 implies f(6)=9 vee f(6)=0$$
Is this correct? Also, how do I rule out $f(6)=0$ since my source only gives $9$ as the answer. I found this post after writing mine but I still don't think that $f(6)=0$ can be ruled out.






integration derivatives definite-integrals continuity







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  • $begingroup$
    $f(x)equiv 0$ is a solution, so you cannot rule out $f(6)=0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 7:33










  • $begingroup$
    See the comment sections of the linked post that discuss the general solution.
    $endgroup$
    – StubbornAtom
    Feb 1 at 10:09
















3












$begingroup$


Let $f:[0,infty)to[0,infty)$ be continuous on $[0,infty)$ and differentiable on $(0,infty)$. If $f(x)=int_{0}^{x}sqrt {f(t)}dt$, then find $f(6)$.
$$f(x)=int_{0}^{x}sqrt {f(t)}dt$$
$$g(x):=sqrt {f(x)}implies(g(x))^2=f(x)=int_{0}^{x}g(t)dtimplies2g(x)g'(x)=g(x)$$
[by FTC-1]
$$implies g=0 vee g'(x)=frac{1}{2} $$
$$(g(0))^2=f(0)=0implies g(x)=frac{x}{2}vee g=0 implies f(6)=9 vee f(6)=0$$
Is this correct? Also, how do I rule out $f(6)=0$ since my source only gives $9$ as the answer. I found this post after writing mine but I still don't think that $f(6)=0$ can be ruled out.






integration derivatives definite-integrals continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f(x)equiv 0$ is a solution, so you cannot rule out $f(6)=0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 7:33










  • $begingroup$
    See the comment sections of the linked post that discuss the general solution.
    $endgroup$
    – StubbornAtom
    Feb 1 at 10:09














3












3








3


2



$begingroup$


Let $f:[0,infty)to[0,infty)$ be continuous on $[0,infty)$ and differentiable on $(0,infty)$. If $f(x)=int_{0}^{x}sqrt {f(t)}dt$, then find $f(6)$.
$$f(x)=int_{0}^{x}sqrt {f(t)}dt$$
$$g(x):=sqrt {f(x)}implies(g(x))^2=f(x)=int_{0}^{x}g(t)dtimplies2g(x)g'(x)=g(x)$$
[by FTC-1]
$$implies g=0 vee g'(x)=frac{1}{2} $$
$$(g(0))^2=f(0)=0implies g(x)=frac{x}{2}vee g=0 implies f(6)=9 vee f(6)=0$$
Is this correct? Also, how do I rule out $f(6)=0$ since my source only gives $9$ as the answer. I found this post after writing mine but I still don't think that $f(6)=0$ can be ruled out.






integration derivatives definite-integrals continuity







share|cite|improve this question









$endgroup$




Let $f:[0,infty)to[0,infty)$ be continuous on $[0,infty)$ and differentiable on $(0,infty)$. If $f(x)=int_{0}^{x}sqrt {f(t)}dt$, then find $f(6)$.
$$f(x)=int_{0}^{x}sqrt {f(t)}dt$$
$$g(x):=sqrt {f(x)}implies(g(x))^2=f(x)=int_{0}^{x}g(t)dtimplies2g(x)g'(x)=g(x)$$
[by FTC-1]
$$implies g=0 vee g'(x)=frac{1}{2} $$
$$(g(0))^2=f(0)=0implies g(x)=frac{x}{2}vee g=0 implies f(6)=9 vee f(6)=0$$
Is this correct? Also, how do I rule out $f(6)=0$ since my source only gives $9$ as the answer. I found this post after writing mine but I still don't think that $f(6)=0$ can be ruled out.






integration derivatives definite-integrals continuity




integration derivatives definite-integrals continuity






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asked Feb 1 at 7:03









RhaldrynRhaldryn

352416




352416












  • $begingroup$
    $f(x)equiv 0$ is a solution, so you cannot rule out $f(6)=0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 7:33










  • $begingroup$
    See the comment sections of the linked post that discuss the general solution.
    $endgroup$
    – StubbornAtom
    Feb 1 at 10:09


















  • $begingroup$
    $f(x)equiv 0$ is a solution, so you cannot rule out $f(6)=0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 7:33










  • $begingroup$
    See the comment sections of the linked post that discuss the general solution.
    $endgroup$
    – StubbornAtom
    Feb 1 at 10:09
















$begingroup$
$f(x)equiv 0$ is a solution, so you cannot rule out $f(6)=0$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 7:33




$begingroup$
$f(x)equiv 0$ is a solution, so you cannot rule out $f(6)=0$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 7:33












$begingroup$
See the comment sections of the linked post that discuss the general solution.
$endgroup$
– StubbornAtom
Feb 1 at 10:09




$begingroup$
See the comment sections of the linked post that discuss the general solution.
$endgroup$
– StubbornAtom
Feb 1 at 10:09










2 Answers
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$begingroup$

$f(x)=frac {x^{2}} 4$ and $f(x)=0$ both satisfy the given equation so $f(6)$ is not uniquely determined.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If you differentiate the equality you get $f'(x)=sqrt{f(x)}$. Now you can solve this separable ODE (note that $f(0)=0$) and get your answer.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

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      2 Answers
      2






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      1












      $begingroup$

      $f(x)=frac {x^{2}} 4$ and $f(x)=0$ both satisfy the given equation so $f(6)$ is not uniquely determined.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $f(x)=frac {x^{2}} 4$ and $f(x)=0$ both satisfy the given equation so $f(6)$ is not uniquely determined.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $f(x)=frac {x^{2}} 4$ and $f(x)=0$ both satisfy the given equation so $f(6)$ is not uniquely determined.






          share|cite|improve this answer









          $endgroup$



          $f(x)=frac {x^{2}} 4$ and $f(x)=0$ both satisfy the given equation so $f(6)$ is not uniquely determined.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 1 at 7:35









          Kavi Rama MurthyKavi Rama Murthy

          73.8k53170




          73.8k53170























              0












              $begingroup$

              If you differentiate the equality you get $f'(x)=sqrt{f(x)}$. Now you can solve this separable ODE (note that $f(0)=0$) and get your answer.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If you differentiate the equality you get $f'(x)=sqrt{f(x)}$. Now you can solve this separable ODE (note that $f(0)=0$) and get your answer.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If you differentiate the equality you get $f'(x)=sqrt{f(x)}$. Now you can solve this separable ODE (note that $f(0)=0$) and get your answer.






                  share|cite|improve this answer









                  $endgroup$



                  If you differentiate the equality you get $f'(x)=sqrt{f(x)}$. Now you can solve this separable ODE (note that $f(0)=0$) and get your answer.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 8:03









                  PierreCarrePierreCarre

                  2,067213




                  2,067213






























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