解析:因为f(-2)=(-2)2-(-2)=6,f(-1)=(-1)2-2×(-1)=3,f(0)=02-2×0=0,f(1)=12-2×1=-1.所以f(x)的值域为{6,3,0,-1}.

答案:{6,3,0,-1}

7.若函数f(x)满足f(2x-1)=x+1,则f(3)=　　　　　.

解析:令2x-1=3,则x=2,故f(3)=2+1=3.

答案:3

8.若函数f(x)=ax2-1,a为正常数,且f(f(-1))=-1,则a的值是　　　　　.

解析:∵f(-1)=a·(-1)2-1=a-1,f(f(-1))=a·(a-1)2-1=a3-2a2+a-1=-1.∴a3-2a2+a=0,∴a=1或a=0(舍去).故a=1.

答案:1

9.求函数y=√(x+2)/(√(6"-" 2x) "-" 1)的定义域,并用区间表示.

解要使函数有意义,则{■(x+2≥0"," @6"-" 2x≥0"," @6"-" 2x≠1"," )┤解得{■(x≥"-" 2"," @x≤3"," @x≠5/2 "," )┤

　　即-2≤x≤3,且x≠5/2.

　　故函数的定义域为{x├|"-" 2≤x≤3",且" x≠5/2┤},

　　用区间表示为["-" 2"," 5/2)∪(5/2 "," 3].

10.已知函数f(x)=(1+x^2)/(1"-" x^2 ).

(1)求f(x)的定义域;

(2)若f(a)=2,求a的值;

(3)求证:f(1/x)=-f(x).