设函数f(x)=x^x+bx+c,A={x|f(x)=x},B={x|f(x-1)=x+1},若A={2},求集合B
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设函数f(x)=x^x+bx+c,A={x|f(x)=x},B={x|f(x-1)=x+1},若A={2},求集合B
![设函数f(x)=x^x+bx+c,A={x|f(x)=x},B={x|f(x-1)=x+1},若A={2},求集合B](/uploads/image/z/18060525-45-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dx%5Ex%2Bbx%2Bc%2CA%3D%7Bx%7Cf%28x%29%3Dx%7D%2CB%3D%7Bx%7Cf%28x-1%29%3Dx%2B1%7D%2C%E8%8B%A5A%3D%7B2%7D%2C%E6%B1%82%E9%9B%86%E5%90%88B)
f(x)=x^2+bx+c
f(x)=x
=>x^2+(b-1)x+c = 0
A={2}
=> 4+2(b-1)+c =0
c = -2-2b ---(1)
for double roots(a)
=>△ =0
(b-1)^2-4c=0
(b-1)^2 +8(b+1) =0
b^2+6b+9=0
(b+3)^2=0
b= -3
c = 4
f(x)= x^2-3x+4
for B,f(x-1)=x+1
=> (x-1)^2 -3(x-1)+4 = x+1
x^2-6x+9=0
(x-3)^2=0
x=3
=> B ={3} #
f(x)=x
=>x^2+(b-1)x+c = 0
A={2}
=> 4+2(b-1)+c =0
c = -2-2b ---(1)
for double roots(a)
=>△ =0
(b-1)^2-4c=0
(b-1)^2 +8(b+1) =0
b^2+6b+9=0
(b+3)^2=0
b= -3
c = 4
f(x)= x^2-3x+4
for B,f(x-1)=x+1
=> (x-1)^2 -3(x-1)+4 = x+1
x^2-6x+9=0
(x-3)^2=0
x=3
=> B ={3} #
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