Question

已知f（x）是一次函数，且满足3f（x+1）-2f（x-1）=2x+17,求f(x)的解析式

写过程

Answer 1

设f(x)=ax+b
3f(x+1)-2f(x-1)
=3[a(x+1)+b]-2[a(x-1)+b]
=2x+17
即3ax+3a+3b-2ax+2a-2b=2x+17
(a-2)x+5a+b-17=0
所以a-2=0 a=2
5a+b-17=0 解得b=7
所以f(x)=2x+7

Answer 2

设f(x)=kx+b
3f(x+1)-2f(x-1)
=3kx+3k+3b-2kx+2k-2b
=kx+5k+b
因为3f(x+1)-2f(x-1)=2x+17
所以kx+5k+b=2x+17
所以k=2
所以5k+b=17
5*2+b=17
b=7
所以f(x)=2x+7

Answer 3

设f(x)=ax+b
根据题意
3[a(x+1)+b]-2[a(x-1)+b]=2x+17
——>3ax+3+3b-2ax+2-2b=2x+17
——>ax+b+5=2x+17
——>a=2，b+5=17
∴f(x)=2x+12

Answer 4

f(x)=ax+b
3f(x+1)-2f(x-1)
=3ax+3+3b-2ax+2-2b
=ax+5+b
a=2
5+b=17
b=12
f(x)=2x+12

Answer 5

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