已知函数f(x)=x的平方+ax+b,且集合A={x x=f(x)}B={x x=f(f(x))}.求证1.A属于B2.A={-1,3}时,用列举法表示B

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已知函数f(x)=x的平方+ax+b,且集合A={x x=f(x)}B={x x=f(f(x))}.求证1.A属于B2.A={-1,3}时,用列举法表示B

f(x)=x^2+ax+bA={x| x=f(x)},B={x| x=f(f(x))}for all x ∈ A=> x = f(x)=> f(x) = f(f(x))=> x = f(f(x)) ( x= f(x))=> x ∈ B=> A is subset of BA={-1,3}x= x^+ax+bx^2+(a-1)x+b=0sum of roots -(a-1) = 2a =...