The exercise [1, p. 86 ex. 3.16] asks to prove that if the eigenfunctions and eigenvalues of an operator \widehat{A} are \{\phi_{n}\} and \{a_{n}\}, respectively (\ensuremath{\widehat{A}}\phi_{n}=a_{n}\phi_{n}) then the eigenfunctions of a function f(x) having an expansion of the form:
f(x)=\sum\limits^{\infty}_{l=0}b_{l}x^{l} (1)

will also be \phi_n  with corresponding eigenvalues f(a_n) , n=0,1... . That is f(\widehat{A})\phi_n=f(a_n)\phi_n.
Solution:To prove this feature, one must have the distributive property of operators in mind which can be stated as:  \widehat{A}^{l}=\widehat{A}^{l-1}\cdot \widehat{A}. Iterative application of the operator \widehat{A}^{l} to the eigenfunction \phi_n gives the following result:
\widehat{A}^{l}\phi_n=\widehat{A}^{l-1}\widehat{A}\phi_n = a_n\widehat{A}^{l-1}\phi_n =...= a^{k}\widehat{A}^{l-k}\phi_n=...=a^{l}_{n}\phi_n. (2)

The function f(\widehat{A}) can be written with the use of the given expansion as: f(\widehat{A})=\sum^{\infty}_{l=0}b_{l}\widehat{A}^{l} Applying the right hand side to an eigenfunction \phi_{n} results in:
f(\widehat{A})\phi_n=\sum\limits^{\infty}_{l=0}b_{l}\widehat{A}^{l}\phi_n
=\sum\limits^{\infty}_{l=0}b_{l}a^{l}_{n}\phi_n
=f(a_n)\phi_n (3)

The last result (3) means that the eigenfunctions of an operator f(\widehat{A}) are the same as for the operator \widehat{A} but the eigenvalues are given by f(a_n) where a_n are the eigenvalues of the operator \widehat{A}; Q.E.D..

[1] Richard L. Liboff: “Introductory Quantum Mechanics”, 2nd edition, ISBN: 0-201-54715-5.