作业帮已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}..已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}.(1)求证:A B;(2)若A={-1,3}时,求集合B.
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![已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}..已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}.(1)求证:A B;(2)若A={-1,3}时,求集合B.](/uploads/image/z/8896573-37-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dx2%2Bax%2Bb%2Ca%2Cb%E2%88%88R%2C%E4%B8%94A%3D%7Bx%7Cx%3Df%28x%29%7D%2CB%3D%7Bx%7Cx%3Df%5Bf%28x%29%5D%7D..%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dx2%2Bax%2Bb%2Ca%2Cb%E2%88%88R%2C%E4%B8%94A%3D%7Bx%7Cx%3Df%28x%29%7D%2CB%3D%7Bx%7Cx%3Df%5Bf%28x%29%5D%7D.%281%29%E6%B1%82%E8%AF%81%3AA+B%3B%282%29%E8%8B%A5A%3D%7B-1%2C3%7D%E6%97%B6%2C%E6%B1%82%E9%9B%86%E5%90%88B.)
已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}..已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}.(1)求证:A B;(2)若A={-1,3}时,求集合B.
已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}.
.已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}.
(1)求证:A B;
(2)若A={-1,3}时,求集合B.
已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}..已知函数f（x）=x2+ax+b,a,b∈R,且A={x|x=f(x)},B={x|x=f[f(x)]}.(1)求证:A B;(2)若A={-1,3}时,求集合B.
f（x）=x^2+ax+b
A=｛x| x=f(x)｝, B={x| x=f(f(x))}
x ∈ A
=> x = f(x)
=> f(x) = f(f(x))
=> x = f(f(x)) ( x= f(x))
=> x ∈ B
=> A is subset of B
A={-1,3}
x= x^+ax+b
x^2+(a-1)x+b=0
sum of roots
-(a-1) = 2
a = -1
product of roots
b=-3
=> f(x) = x^2-x-3
f(f(x)) = (x^2-x-3)^2 -(x^2-x-3) -3 = x
(x^2-x)^2-6(x^2-x) + 9 - x^2=0
x^4-2x^3-6x^2+6x+9=0
(x+1)(x-3)(x^2-3)=0
x = -1 or 3 or -√3 or √3
B = {-1 , 3 , -√3 , √3}