Question

证明函数f(x)=x/x2+1在(0,1)上是增函数

Answer 1

f(x)=x/(x²+1)
x∈（0，1)
令0＜x1＜x2＜1
f(x2)-f(x1)
= x2/(x2²+1) - x1/(x1²+1)
= [ x2(x1²+1) - x1(x2²+1) ] / [(x1²+1)(x2²+1)]
= [(x1² x2+ x2 - x1x2²-x1 ] / [(x1²+1)(x2²+1)]
= [(x1² x2-x1) - (x1x2² - x2)] / [(x1²+1)(x2²+1)]
= [x1(x1x2-1) - x2 (x1x2-1)] / [(x1²+1)(x2²+1)]
= [(x1x2-1)(x1- x2)] / [(x1²+1)(x2²+1)]
∵0＜x1＜x2＜1
∴x1x2-1＜0；x1- x2＜0；(x1²+1)(x2²+1)＞0
∴ [(x1x2-1)(x1- x2)] / [(x1²+1)(x2²+1)]＞0
∴f(x2)＞f(x1)，得证。

Answer 2

f(x)=x/(x^2+1)
f'(x) =(x^2+1- 2x^2 )/(x^2+1)^2
= (-x^2+1)/(x^2+1)^2 >0 ;x 在(0,1)

=>f(x)=x/(x^2+1) 在(0,1)上是增函数