题目

已知函数f(x)=1-2/(a^x+1) (a大于0且不等于1)
1 若g(x）=f(x-0.5)+1 证明g(x)+g(1-x)是一个常数
2在1的条件下,求g(0）+g(0.1)+g(0.2)+g(0.3)+.+g(0.9)+g(1)的值

提问时间：2021-04-20

答案

f(x)=1-2/(a^x+1)=(a^x-1)/(a^x+1)
g(x)=f(x-0.5)+1
g(x)+g(1-x)
=f(x-0.5)+f(0.5-x)+2
=[a^(x-0.5)-1]/[a^(x-0.5)+1]+[a^(0.5-x)-1]/[a^(0.5-x)+1]+2
=[a^x/√a-1]/[a^x/√a+1]+[√a/a^x-1]/[√a/a^x+1]+2
=[a^x-√a]/[a^x+√a]+[√a-a^x]/[√a+a^x]+2
=0+2
=2
g(0)+g(0.1)+g(0.2)+...+g(0.9)+g(1)
=[g(0)+g(1)]+[g(0.1)+g(0.9)]+...+[g(0.4)+g(0.6)]+[g(0.5)+g(0.5)]/2
=2+2+...+2+2/2
=5*2+1
=11