设函数 f(x)=x²+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)=x²+bx+c,A=﹛x |f(x)=x﹜,B=﹛x |f(x-1)=x+1﹜,若A=﹛2﹜,求集合B](/uploads/image/z/6926143-31-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0+f%EF%BC%88x%EF%BC%89%3Dx%26%23178%3B%2Bbx%2Bc%2CA%3D%EF%B9%9Bx+%7Cf%EF%BC%88x%EF%BC%89%3Dx%EF%B9%9C%2CB%3D%EF%B9%9Bx+%7Cf%EF%BC%88x-1%EF%BC%89%3Dx%2B1%EF%B9%9C%2C%E8%8B%A5A%3D%EF%B9%9B2%EF%B9%9C%2C%E6%B1%82%E9%9B%86%E5%90%88B)
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设函数 f(x)=x²+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)=x²+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)=x²+bx+c,A=﹛x |f(x)=x﹜,B=﹛x |f(x-1)=x+1﹜,若A=﹛2﹜,求集合B
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}
B={2,5}
A={xIf(x)=x},得到x^2+bx+c=x,所以x^2+(b-1)x+c=0,因为A=﹛2﹜,所以
2*2=-(b-1),2+2=c解得c=4,b=-3;所以
B=﹛x |f(x-1)=x+1﹜得到(x-1)^2-3(x-1)+4=x+1,解得x=3±根号2,所以
集合B={3+根号2,3-根号2}