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)的最小值级对应的x值

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

Answer 2

显然f(0)=f(1)=b
则log2 a=0或1
而由于a不等于1,则只能由a=2来满足log2 a=1
则f(a)=f(2)=b+2
log2[f(a)]=log2[b+2],则log2[b+2]=2
b+2=2^2=4
b=2

f(x)=x的平方-x+2
=(x-1/2)^2 +7/4
那么f(log2 x)=(log2 x-1/2)^2 +7/4
则当log2 x =1/2时有极小值,此时x=2^(1/2)即√2;
极小值为7/4