Question

已知f（x）是一次函数，且满足3f（x+1）-2f（x-1）=2x+17，则f（x）=

Answer 1

设：f（x）=kx+b
3f（x+1）-2f（x-1）=2x+17
则有：
当x=0时有：
3k+3b+2k-2b=17..................1
当x=1时有：
6k+3b-2b=19.......................2
联立1、2得：
k=2,b=7所以：f(x)=2x+7

Answer 2

设f(x)=ax+b
3f（x+1）-2f（x-1）=ax+5a+b=2x+17
得a=2, b=7
f(x)=2x+7

Answer 3

令f(x) = ax + b
3f（x+1)-2f(x-1）= 3(a(x+1)+b) - 2(a(x-1)+b) = 2x + 17
3ax + 3a + b - 2ax +2a - 2b = 2x + 17
ax + 5a-b = 2x + 17
a = 2, 5a-b = 17
b = -7
f(x) = 2x - 7

Answer 4

解 设f（x)=ax+b 则 3f(x+1）-2f(x-1)=3(ax+a+b)-2(ax-a+b)=ax+5a+b=2x+17
所以 a=2
5a+b=17 b=7
所以f(x)=2x+7

Answer 5

设：f(x)=ax+b
则：3f（x+1）-2f（x-1）=3ax+3a+3b-(2ax-2a+2b)=ax+5a+b=2x+17
a=2
5a+b=17, b=7

f(x)=2x+7