若2≤x≤8,求函数y=(log1/4x)^2+log1/4x^2+5的最值

由题意y=(log1/4x)^2+log1/4x^2+5
=(log1/4x)^2+2log1/4x+5
=[log1/4(x)+1]^2+4
因为log1/4(x)在定义域内递减,又因为2≤x≤8,
所以log1/4(8)=-3/2≤log1/4(x)≤log1/4(2)=-1/2
即-1/2≤log1/4(x)+1≤1/2
所以0≤[log1/4(x)+1]^2≤1/4
得出4≤[log1/4(x)+1]^2+4≤17/4
所以函数最小值为4,最大值为17/4