已知f(x)=log2(1+x)+ log2(1-x) f[(a+b)/(1+ab)]=2 f(-b)=1/2 求f(a)=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 07:45:34
已知f(x)=log2(1+x)+ log2(1-x) f[(a+b)/(1+ab)]=2 f(-b)=1/2 求f(a)=?
xRN0~mk[$ a %RTu)l `@BꊀT)iŔWwN ҩ|aU/Y nGQYrhȘL $c0=V[JRynχc5x-FPT2lTxbbǒ%O*ׇ8Vۭfj~jl%)h{{K ^/F貧`>_蹈(rzHHo*a;}F#xx'A]uy~[p{$@y'l$C\:a)

已知f(x)=log2(1+x)+ log2(1-x) f[(a+b)/(1+ab)]=2 f(-b)=1/2 求f(a)=?
已知f(x)=log2(1+x)+ log2(1-x) f[(a+b)/(1+ab)]=2 f(-b)=1/2 求f(a)=?

已知f(x)=log2(1+x)+ log2(1-x) f[(a+b)/(1+ab)]=2 f(-b)=1/2 求f(a)=?
题目应为:f(x) = log2(1+x) - log2(1-x)
f(x) = log2(1+x) - log2(1-x) = log2 [(1+x)/(1-x) ]
f(-x) = log2 [(1-x)/(1+x) ] = -l og2 [(1+x)/(1-x) ] = -f(x),奇函数
f(a) = log2 [(1+a)/(1-a) ]
f(b) = log2 [(1+b)/(1-b) ]
f[(a+b)/(1+ab)]=2
log2【{1+(a+b)/(1+ab)} / {1-(a+b)/(1+ab)} 】= 2
log2【(1+ab+a+b) / (1+ab-a-b) 】 = 2
log2【(1+a)(1+b)/[(1-a)(1-b)] 】= 2
log2【(1+a)/(1-a) * (1+b)/(1-b) 】= 2
log2【(1+a)/(1-a)】+ log2【(1+b)/(1-b) 】= 2
f(a) + f(b) = 2 .(1)
f(-b)= 1/2
f(b) = -f(-b) = -1/2 .(2)
f(a) = 2-f(b) = 2-(-1/2) = 5/2