Browse · MathNet Print → China Mathematical Competition China precalculus Problem Suppose that f(x)=1+x2x and f(n)(x)=f[f[f…f(x)]]. Then f(99)(1)=. Solution — click to reveal We have f(1)(x)f(2)(x)f(99)(x)=f(x)=1+x2x,=f[f(x)]=1+2x2x,⋮=1+99x2x. Therefore, f(99)(1)=101. Final answer 1/10 Techniques Functions ← Previous problem Next problem →