已知函数f(x)=ex+1ex-1.(Ⅰ)求f(x)的反函数f-1(x);(Ⅱ)讨论f(x)的奇偶性.-数学

题目简介

已知函数f(x)=ex+1ex-1.(Ⅰ)求f(x)的反函数f-1(x);(Ⅱ)讨论f(x)的奇偶性.-数学

题目详情

已知函数f(x)=
ex+1
ex-1

(Ⅰ)求f(x)的反函数f-1(x);
(Ⅱ)讨论f(x)的奇偶性.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由y=
ex+1
ex-1
,得yex-y=ex+1,
从而yex-ex=y+1,(y-1)ex=y+1,∴ex=class="stub"y+1
y-1

ex=class="stub"y+1
y-1
>0
,得y<-1,或y>1.
再由ex=class="stub"y+1
y-1
,得x=lnclass="stub"y+1
y-1
(y<-1,或y>1)

f-1(x)=lnclass="stub"x+1
x-1
(x<-1或x>1).
(Ⅱ)f(x)=
ex+1
ex-1
中,∵ex-1≠0,∴x≠0.∴函数f(x)的定义域为{x|x≠0},它关于原点对称.
f(-x)=
e-x+1
e-x-1
=
(e-x+1)•ex
(e-x-1)•ex
=
1+ex
1-ex
=-
ex+1
ex-1
=-f(x)

∴函数f(x)是奇函数.

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