Inverse of function and derivative












0












$begingroup$


The problem says:



If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.



I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.



The answer says (f^-1)'(2)=1/4



Thanks






calculus inverse







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    0












    $begingroup$


    The problem says:



    If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.



    I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.



    The answer says (f^-1)'(2)=1/4



    Thanks






    calculus inverse







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      The problem says:



      If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.



      I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.



      The answer says (f^-1)'(2)=1/4



      Thanks






      calculus inverse







      share|cite|improve this question











      $endgroup$




      The problem says:



      If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.



      I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.



      The answer says (f^-1)'(2)=1/4



      Thanks






      calculus inverse




      calculus inverse






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 13:10







      L. G.

















      asked Dec 15 '18 at 12:37









      L. G. L. G.

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          $begingroup$

          Hint: Since $f(1)=2$, $f^{-1}(2)=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ok, but how did you solve 2=f(x) for x?
            $endgroup$
            – L. G.
            Dec 15 '18 at 12:47










          • $begingroup$
            $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 12:49











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          1






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          active

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          $begingroup$

          Hint: Since $f(1)=2$, $f^{-1}(2)=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ok, but how did you solve 2=f(x) for x?
            $endgroup$
            – L. G.
            Dec 15 '18 at 12:47










          • $begingroup$
            $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 12:49
















          1












          $begingroup$

          Hint: Since $f(1)=2$, $f^{-1}(2)=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ok, but how did you solve 2=f(x) for x?
            $endgroup$
            – L. G.
            Dec 15 '18 at 12:47










          • $begingroup$
            $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 12:49














          1












          1








          1





          $begingroup$

          Hint: Since $f(1)=2$, $f^{-1}(2)=1$.






          share|cite|improve this answer









          $endgroup$



          Hint: Since $f(1)=2$, $f^{-1}(2)=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 12:45









          José Carlos SantosJosé Carlos Santos

          160k22127232




          160k22127232












          • $begingroup$
            ok, but how did you solve 2=f(x) for x?
            $endgroup$
            – L. G.
            Dec 15 '18 at 12:47










          • $begingroup$
            $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 12:49


















          • $begingroup$
            ok, but how did you solve 2=f(x) for x?
            $endgroup$
            – L. G.
            Dec 15 '18 at 12:47










          • $begingroup$
            $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 12:49
















          $begingroup$
          ok, but how did you solve 2=f(x) for x?
          $endgroup$
          – L. G.
          Dec 15 '18 at 12:47




          $begingroup$
          ok, but how did you solve 2=f(x) for x?
          $endgroup$
          – L. G.
          Dec 15 '18 at 12:47












          $begingroup$
          $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
          $endgroup$
          – José Carlos Santos
          Dec 15 '18 at 12:49




          $begingroup$
          $$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
          $endgroup$
          – José Carlos Santos
          Dec 15 '18 at 12:49


















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