已知函数f(x)=log2(x+1).(1)若2f(x)小于等于f(6x-3),求x的取值集合D
1个回答

1)2f(x)≤f(6x-3)

2log2(x+1)≤log2(6x-3+1)

得: (x+1)^2≤(6x-2)

x^2+2x+1-6x+2≤0

x^2-4x+3≤0

(x-1)(x-3)≤0

1≤x≤3

经检验它满足不等式

则D={x|1≤x≤3}

2)F(x)=[log2(x)]^2-log2(x^2)=[log2(x)]^2-2log2(x)=[log2(x)-1]^2-1

当1≤x≤3时,0≤log2(x)≤log2(3),

当log2(x)=1时,即x=2时,F(x)取最小值-1

当log2(x)=0时,即x=1时,F(x)取最大值0

所以F(x)的值域为[-1,0]