Question

设f（x=1）=x²+3x+5，求f（x），f（x-1）

Answer 1

1、若题目为：已知f(x+2)=x^2+3x+5,求f(x),f(x-1)
解：令x+2=t x=t-2
f(x+2)=f(t)=(t-2)²+3(t-2)+5
= t²-t+3
f(t)= t²-t+3
所以f（x）= x²-x+3
f(x-1)=(x-1)²-(x-1)+3=x²-3x+4
所以f（x）= x²-x+3
f(x-1)=x²-3x+4

2、若题目为：已知f(x+1)=x^2+3x+5,求f(x),f(x-1)
解：令x+1=t x=t-1
f(x+1)=f(t)=(t-1)²+3(t-1)+5
= t²+t+3
f(t)= t²+t+3
所以f（x）= x²+x+3
f(x-1)=(x-1)²+(x-1)+3=x²-x+3
综上， f（x）= x²+x+3
f(x-1)=x²-x+3

Answer 2

感觉条件有问题啊 应该是f(x+1)。f(x)=x平方+x+3 f(-)=x平方-x+3

Answer 3

f(x+1) = x²+3x+5 = x²+2x+1+x+1+3 = (x+1)² + (x+1) + 3
f(x) = x²+x+3
f(x-1) = (x-1)²+(x-1)+3 = x²-x+3

Answer 4

f(x+1)=(x+1)^2+(x+1)+3
f(x)=x²+x+3
f(x-1)=(x-1)^2+(x-1)+3=x²-x+3