因f(x)=log2(x/4)log2(2x)=[log2(x)-2][1+log2(x)]=[log2(x)]^2-log2(x)-2=[log2(x)-1/2]^2-9/4而1≤x≤8,即0≤log2(x)≤3则-9/4≤f(x)≤4令g(x)=log2(x),显然g(x)为增函数,且当1≤x≤8时,0≤g(x)≤3令h(x)=(x-1/2)^...