Question

已知2根号2<=x<=8,求函数f(x)=log2x/2log2x/4的最大值与最小值

Answer 1

f(x)=log2 x/2log2 (x/4)=log2 x/2(log2 x-log2 4)=log2 x/2(log2 x-2)由于2√2=√8=2^(3/2) [2的二分之三次方] 8=2^3 [2的三次方]于是2√2<=x<=8 即2^2(3/2) ≤ x ≤2^3于是取以2为底的对数可得，3/2 ≤ log2 x ≤ 3那么log2 x的下限是3/2，即当x=2√2时，f(x)=(3/2)/2(3/2-2)= -3/2当log2 x达到上限3时，即当x=8时，f(x)=3/2(3-2)=3/2所以，原函数的最小值为-3/2，最大值为3/2