已知f(x)=x²-x+k,且log₂f(a)=k,a>0且a≠1.(1)求a,k的值; (2)当x为何值时,f(log‹a› x)
有最小值,并求出最小值 .
(1)∵log₂f(a)=log₂(a²-a+k)=k,∴a²-a+k=2^k,可取k=2,得a²-a-2=(a-2)(a+1)=0,故a=2.
(2).f(log₂x)=[log₂x]²-log₂x+2=u²-u+2=(u-1/2)²-1/4+2=(u-1/2)²+7/4
其中u=log₂x,当u=2时f(log₂x)有最小值7/4.此时x=4.