> 数学 >
已知函数f(x)=
10x10−x
10x+10−x
,判断f(x)的奇偶性和单调性.
人气:182 ℃ 时间:2019-08-18 11:29:41
解答
(1)已知函数f(x)=
10x10−x
10x+10−x
=
102x−1
102x+1
,x∈R

f(x)=
10−x10x
10−x+10x
=
102x−1
102x+1
=−f(x),x∈R

∴f(x)是奇函数
(2)f(x)=
102x−1
102x+1
,x∈R
,设x1,x2∈(-∞,+∞),且x1<x2
f(x1) −f(x2) =
102x1−1
102x1+1
102x2−1
102x2+1
=
2(102x1102x2)
(102x1+1)(102x2+1)
2(100x1100x2)
(102x1+1)(102x2+1)

因为x1<x2,所以100x1100x2,所以f(x1)-f(x2)<0,即f(x1)<f(x2),
∴f(x)为增函数.
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版