Question

设f(x)=x方+bx+c.集合A={x|f(x)=x}.B={x|f(x-1)=x+1}若A=2,求集合B.

Answer 1

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)

=>△ =0

f(x)= x^2-3x+4

for B,f(x-1)=x+1

=> (x-1)^2 -3(x-1)+4 = x+1

x=3

=> B ={3}