Question

已知函数f(x+1)=x^2-3x+2 (1)求f(2)和f(a)的值(2)求f(x)和f(x-1)

Answer 1

(1)、令x+1=2，则x=1
∴f(2)=1²-3*1+2=0
令x+1=a，则x=a-1
∴f(a)=(a-1)²-3(a-1)+2
=(a-1-1)(a-1-2)
=(a-2)(a-3)
=a²-5a+6

(2)、f(x+1)=x²-3x+2
=(x²+2x+1)-5(x+1)+6
=(x+1)²-5(x+1)+6
∴f(x)=x²-5x+6
f(x-1)=(x-1)²-5(x-1)+6
=x²-7x+12

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Answer 2

令x=1代入得f(2)=1-3+2=0;
令t=x+1,则x=t-1,即f(t)=(t-1)^2-3(t-1)+2=t^2-5t+6,
令t=x即f(x)=x^2-5x+6
将x=a代入得f(a)=a^2-5a+6
令t=x-1,则f(x-1)=(x-1)^2-5(x-1)+6=x^2-7x+12

Answer 3

f（2）=f(1+1) =1^2-3+2=0
f(x+1)=(x+1)^2-5(x+1)+6
f(x)=x^2-5x+6
f(x-1)=(x-1)^2-5(x-1)+6=x^2-7x+12

Answer 4

f(x+1)=x^2-3x+2
f(x)=(x-1)^2-3(x-1)+2
=x^2-2x+1-3x+3+2
=x^2-5x+6

f(2)=2^2-5*2+6=0
f(a)=a^2-5a+6

f(x-1)=(x-1)^2-5(x-1)+6
=x^2-2x+1-5x+5+6
=x^2-7x+12