已知函数f(x+1)=x的平方-3x+2,求f(x)值,和f(x-1)的值
问题描述：
已知函数f(x+1)=x的平方-3x+2,求f(x)值,和f(x-1)的值
答
f(x+1) = x^2 - 3x + 2
设M = X + 1,则X = M - 1.代入得:
f(M) = (M - 1)^2 - 3(M - 1) + 2 = M^2 - 5M + 6
所以,该函数就是:f(x) = x^2 - 5x + 6
则f(x-1) = (x - 1)^2 - 5(x - 1) + 4 = x^2 - 7x + 12
即 f(x-1)= x^2 - 7x + 12
答
先代x=x-1，得f(x)=(x-1)^2-3(x-1)+2,即f(x)=x^2-5x+6
f(x-1)=x^2-7x+12
答
令a=x+1
则x=a-1
所以f(a)=(a-1)²-3(a-1)+2=a²-5a+6
所以f(x)=x²-5x+6
f(x-1)=(x-1)²-5(x-1)+6=x²-7x+12