Question

f（x）=(4^x-1)/(4^x+1),若x1>0,x2>0且f（x1）+f（x2）=1则f（x1+x2）的最小值为

Answer 1

f(x1)+f(x2)=(4^x1-1)/(4^x1+1)+(4^x2-1)/(4^x2+1)=1,

∴(4^x1-1)(4^x2+1)+(4^x2-1)(4^x1+1)=(4^x1+1)(4^x2+1),

∴4^(x1+x2)=4^x1+4^x2+3>=2*2^(x1+x2)+3,

设u=2^(x1+x2)>0,则u^2-2u-3>=0,u>=3,

f(x1+x2)=[4^(x1+x2)-1]/[4^(x1+x2)+1]=1-2/(4^x1+4^x2+4)>=1-2/(2u+4)>=1-2/10=4/5,选B.