設(shè)-1
f(x2)-f(x1)=x2/(1+x2²) -x1/(1+x1²)
=[x2(1+x1²)-x1(1+x2²)]/[(1+x1²)(1+x2²)]
=(x2+x1²x2-x1-x1x2²)/[(1+x1²)(1+x2²)]
=[(x2-x1)-x1x2(x2-x1)]/[(1+x1²)(1+x2²)]
=(x2-x1)(1-x1x2)/[(1+x1²)(1+x2²)]
-1
x2>x1 x2-x1>0
(1+x1²)(1+x2²)>0
(x2-x1)(1-x1x2)/[(1+x1²)(1+x2²)]>0
f(x2)>f(x1),函數(shù)在區(qū)間(-1,1)上單調(diào)遞增.
導(dǎo)數(shù)法：
f'(x)=[x'(1+x²)-x(1+x²)']/(1+x²)²
=(1+x²-2x²)/(1+x²)²
=(1-x²)/(1+x²)²
-1
f'(x)>0,函數(shù)在區(qū)間(-1,1)上單調(diào)遞增.