Question

已知定义域为r的函数f(x)=1-2x/2*x+1是奇函数

判断并证明函数的单调性

Answer 1

f(x) = (1-2^x)/(2^x+1) = {-(2^x+1)+2}/(2^x+1) = -2 + 2/(2^x+1)
2^x单调增
2^x+1单调增
2/(2^x+1)单调减
f(x) = -2 + 2/(2^x+1)为减函数

证明：
令x1＜x2
f(x2) - f(x1)
= {-2 + 2/(2^x2+1)} - {-2 + 2/(2^x1+1)}
= 2/(2^x2+1) - 2/(2^x1+1)
= 2(2^x1+1-2^x2-1)/{(2^x2+1) (2^x1+1)}
= 2(2^x1-2^x2)/{(2^x2+1) (2^x1+1)}
∵2^x2+1 ＞1； 2^x1+1＞1； 2^x1-2^x2＜0
∴f(x2) - f(x1) = 2(2^x1-2^x2)/{(2^x2+1) (2^x1+1)} ＜0
∴f(x2) ＜ f(x1) ，得证。