Question

已知f(x)=(2^x-1)/(2^x+1)为奇函数，用定义法判断f(x)在其定义域上的单调性

Answer 1

f(x)=(2^x-1)/(2^x+1)
先谨烂把f(x)变换形式纳做
f(x)=(2^x-1)/(2^x+1)
=(2^x+1-2)/(2^x+1)
=1-2/(2^x+1)

任取x1<x2∈R
f(x1)-f(x2)
=1-2/(2^x1+1)-(1-2/(2^x2+1))
=2/(2^x2+1)-2/(2^x1+1)
=(2(2^x1+1)-2(2^x2+1))/((2^x1+1)(2^x2+1))
=(2(2^x1-2^x2))/((2^x1+1)(2^x2+1))

因为(2^x1+1)(2^x2+1)>0,2^x1-2^x2<祥茄漏0
所以f(x1)-f(x2)<=0,函数递增

Answer 2

令x2>x1
f(x2)-f(x1)=(2^x2-1)/(2^x2+1)-(2^x1-1)/(2^x1+1)
=[(2^x2-1)(2^x1+1)-(2^x1-1)(2^x2+1)]/[(2^x2+1)(2^x1+1)]
=2(2^x2-2^x1)]/[(2^x2+1)(2^x1+1)]
∵x2>x1
2^x2-2^x1>0
(2^x2+1)(2^x1+1)>0
∴f(x2)-f(x1)>0
f(x)在谨宴其定此正义域上森晌悔的单调增加

Answer 3

f(x)=1-2/(2^x+1)
X1<X2
2^x1+1<2^x2+1
f(x1)<f(x2)
递增