Question

已知函数f（x+1）=x²+x 则f（x）=

已知函数f（x+1）=x²+x 则f（x）=
求解~

Answer 1

记x+1=t得x=t-1，代入得f(t)=(t-1)²+(t-1)=t²-t，从而f(x)=x²-x。

Answer 2

令t=x+1,则x=t-1,f(t)=(t-1)²＋(t-1)=t²-t,所以f(x)=x²-x.

Answer 3

设t=x+1,则x=t-1.
f(t)=（t-1)²+（t-1)=t²-t.
所以f(x)=x²-x.

Answer 4

令x=y-1 带进去就行了f(y)=y方-y
就是f（x）=x方-x

Answer 5

f（x+1）=x²+x=(x+1)²-(x+1) f（x）=x²-x