Answer
$$f\left( x \right) = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right)$$
Work Step by Step
$$\eqalign{
& f''\left( x \right) = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right) \cr
& {\text{Integrate}} \cr
& f'\left( x \right) = \int {f''\left( x \right)} dx \cr
& f'\left( x \right) = \frac{1}{2}\int {\left( {{e^x} + {e^{ - x}}} \right)} dx \cr
& f'\left( x \right) = \frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right) + {C_1}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}f'\left( 0 \right) = 0 \cr
& 0 = \frac{1}{2}\left( {{e^0} - {e^0}} \right) + {C_1} \cr
& {C_1} = 0 \cr
& {\text{Substitute }}{C_1}{\text{ into }}\left( {\bf{1}} \right) \cr
& f'\left( x \right) = \frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right) \cr
& {\text{Integrate}} \cr
& f\left( x \right) = \int {f'\left( x \right)} dx \cr
& f\left( x \right) = \int {\left[ {\frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right)} \right]} dx \cr
& f\left( x \right) = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right) + {C_2}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Use the initial condition }}f\left( 0 \right) = 1 \cr
& 1 = \frac{1}{2}\left( {{e^0} + {e^0}} \right) + {C_2} \cr
& 1 = \frac{1}{2}\left( 2 \right) + {C_2} \cr
& {C_2} = 0 \cr
& {\text{Substitute }}{C_2}{\text{ into }}\left( {\bf{2}} \right) \cr
& f\left( x \right) = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right) \cr} $$
