8.已知f(x)={■(0"," x>0"," @"-" 1"," x=0"," @2x"-" 3"," x<0"," )┤则f(f(f(5)))等于　　　　　.

解析:f(f(f(5)))=f(f(0))=f(-1)=2×(-1)-3=-5.

答案:-5

9.已知函数f(x)={■("-" 2x"," x"∈(-∞,-" 1")," @2"," x"∈[-" 1"," 1" ," @2x"," x"∈(" 1"," +"∞)." )┤

(1)求f("-" 3/2),f(1/2),f(4.5),f(f(1/2));

(2)若f(a)=6,求a的值.

解(1)∵-3/2∈(-∞,-1),

　　∴f("-" 3/2)=-2×("-" 3/2)=3.

　　∵1/2∈[-1,1 ,∴f(1/2)=2.

　　又2∈(1,+∞),∴f(f(1/2))=f(2)=2×2=4.

　　∵4.5∈(1,+∞),∴f(4.5)=2×4.5=9.

　　(2)经观察可知a∉[-1,1 ,否则f(a)=2.

　　若a∈(-∞,-1),令-2a=6,得a=-3,符合题意;

　　若a∈(1,+∞),令2a=6,得a=3,符合题意.

　　故a的值为-3或3.

10.设函数f(x)={■(x^2+bx+c"," x≤0"," @2"," x>0"," )┤若f(-2)=f(0),f(-1)=-3,求关于x的方程f(x)=x的解.

解∵当x≤0时,f(x)=x2+bx+c,∴f(-2)=(-2)2-2b+c,f(0)=c,f(-1)=(-1)2-b+c.

　　∵f(-2)=f(0),f(-1)=-3,

　　∴{■("(-" 2")" ^2 "-" 2b+c=c"," @"(-" 1")" ^2 "-" b+c="-" 3"," )┤解得{■(b=2"," @c="-" 2"." )┤

则f(x)={■(x^2+2x"-" 2"," x≤0"," @2"," x>0"," )┤当x≤0时,由f(x)=x得x2+2x-2=x,得x=-2或x=1.