Question

已知函数f(x)=X/1+lxl,设f1(x)=f(x),fn+1(x)=f【fn(x)】,(n∈N*)

Answer 1

(1) f2(x)=f[f1(x)]=f1(x)/[1+|f1(x)|]=[x/(1+|x|)]/[1+|x/(1+|x|)|]=[x/(1+|x|)]/[1+|x|/(1+|x|)]=x/(1+2|x|)

f3(x)=f[f2(x)]=f2(x)/[1+|f2(x)|]=[x/(1+2|x|)]/[1+|x/(1+2|x|)|]=[x/(1+2|x|)]/[1+|x|/(1+2|x|)]=x/(1+3|x|)

依此猜想 fn(x)=x/(1+n|x|),可用数学归纳法验证（略）

(2)当x≥0时,fn(x)=x/(1+nx),f'n(x)=(1+nx-nx)/(1+nx)²=1/(1+nx)²>0,fn(x)在[0,+∞)上是增函数

又易知fn(x)为奇函数,其图像关于原点对称,所以在(-∞,0]上也是增函数,由于fn(x)在x=0处有定义（连续）,从而fn(x)在R上是增函数.

(3) |fn(x)|=|x|/(1+n|x|)=1/(1/|x| +n)