+26367 If f(x)=x^5-1/3, find f^-1(-31/96).
\(\begin{array}{|rcll|} \hline \boxed{f\left(f^{-1}\left(\dfrac{-31}{96}\right)\right) = \dfrac{-31}{96} \qquad (1) } \\\\ f(x) &=& x^5-\dfrac{1}{3} \quad | \quad x = f^{-1}\left(\dfrac{-31}{96}\right) \\\\ f\Big(f^{-1}\left(\dfrac{-31}{96}\right)\Big) &=& \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5-\dfrac{1}{3} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5-\dfrac{1}{3} = \dfrac{-31}{96} \quad | \quad \text{ see } (1) \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{1}{3} - \dfrac{31}{96} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{96-3\cdot 31}{3\cdot 96} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{3}{3\cdot 96} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{1}{96} \\\\ && f^{-1}\left(\dfrac{-31}{96}\right) = \sqrt[5]{\dfrac{1}{96}} \\\\ && f^{-1}\left(\dfrac{-31}{96}\right) = \sqrt[5]{0.01041666667} \\\\ && \mathbf{ f^{-1}\left(\dfrac{-31}{96}\right) = 0.40137078088 } \\ \hline \end{array}\)
