奇函数f(x)满足f(x+2)=f(x)当x属于(0,1)时f(x)=2^x则f(log0.5底23)=

问题描述:

奇函数f(x)满足f(x+2)=f(x)当x属于(0,1)时f(x)=2^x则f(log0.5底23)=

log0.5底23=lg23/lg0.5=lg23/(-lg2)所以f(log0.5底23)=f[-log2(23)],奇函数,所以=-f[log2(23)]由f(x+2)=f(x),所以f(x+4)=f(x+2+2)=f(x+2)=f(x)所以4是f(x)的周期所以=-f[log2(23)-4]=-f[log2(23)-log2(16)]=-f[log2(...