Tip: As \(f_{n+1}(x)=f_1(f_n(x)) %speech% ,f n plus 1 of x equals f one of f n of x,\), you can prove the value of \(f_{2014}(x)\) by induction.
Tip: For example:
\(f_1(x)=\displaystyle\frac x{x+1} %speech% ,f 1 of x equals x over x plus 1,\), \(f_2(x)=\dfrac {x}{2x+1}\).
