Question

已知函数fx=a2^x+1/2^x+1是奇函数(a为常数)求单调性

不要用导数谢谢还没学

Answer 1

f(x) = {a*2^x + 1} / { 2^x+1}
奇函数
f(-x) = -f(x)
{ a*2^(-x) +1} / {2^(-x)+1 } = - {a*2^x + 1} / { 2^x+1}
左边分子分母同乘以2^x：
{ a +2^x} / {1+2^x } = {-1-a*2^x } / { 1+2^x}
a = -1
f(x) = {-2^x + 1} / { 2^x+1} = {-(2^x+1)+ 2} / { 2^x+1} = -1 + 2/(2^x+1)
2^xd单调增；
2^x+1单调增；
2/(2^x+1)单调减；
-1 + 2/(2^x+1)单调减。
f(x)在（-无穷大，+无穷大）上单调减

Answer 2

解由函数fx=a2^x+1/2^x+1定义域为奇函数，且为奇函数
则f(0)=0
即a×2^0+1=0
解得a=-1
故f(x)=(-2^x+1)/(2^x+1)
由f(x)=(-2^x+1)/(2^x+1)
=(-(2^x+1)+2)/(2^x+1)
=-1+2/(2^x+1)
设x1，x2属于R且x1＜x2
则f(x1)-f(x2)
=[-1+2/(2^x1+1)]-[-1+2/(2^x2+1)]
=2/(2^x1+1)-2/(2^x2+1)
=2(2^x2-2^x1）/(2^x2+1)(2^x1+1)
由x1＜x2
则2^x1＜2^x2
则2^x1-2^x2＜0
即2^x2-2^x1＞0
即2(2^x2-2^x1）/(2^x2+1)(2^x1+1)＞0
故f(x1)-f(x2)＞0
故f(x)在R上为减函数