Question

设函数f(x)=x的平方-x+b且f(log2a)=b,log2[f(a)]=2(a不等于1）,求f(log2x)的最小

Answer 1

∵f(log2a)=(log2[a])^2-log2[a]+b=b

a不等于1

∴log2[a]=1 a=2

∵log2[f(a)]=2

∴f(a)=f(2)=4

∵f(2)=2^2-2+b=4

∴b=2

f(log2x)=[log2x-1/2]^2+7/4

故当log2x=1/2时,f(log2x)min=7/4

此时x=2^(1/2)=√2