1 已知函数f(x)=-x+1,x

1 已知函数f(x)=-x+1,x

挣你这10分真不容易..
1.若x＜0,x+(x+1)(-x)≤1,解得x＜0
若x≥0,x+x(x+1)≤1,同上,解得0≤x≤2½-1
综上,x≤2½-1
2.若log2(x)＞0,原式=log2(x)+logx(2)+1≥1+2(log2(x)*logx(2))½=3
若log2(x)＜0.原式=1-(-log2(x)-logx(2))≤1-2(log2(x)*logx(2))½=0
所以值域为x≤0和x≥3
3.x0∈A,则0≤x0＜1/2,f(f(x0))=f(x0+1/2),x0+1/2∈B,则f(f(x0))=2(1-(x0+1/2))=1-2x0,则1-2x0∈A,0≤1-2x0＜1/2,解得1/4＜x0≤1/2,
综上,1/4＜x0＜1/2