令一次函数f(x)=kx+b,则
f(x+1)=k(x+1)+b=kx+k+b
3f(x+1)=3kx+3k+3b
f(x-1)=k(x-1)+b=kx-k+b
2f(x-1)=2kx-2k+2b
3f(x+1)-2f(x-1)=(3kx+3k+3b)-(2kx-2k+2b)
=3kx+3k+3b-2kx+2k-2b
=kx+5k+b
∵3f(x+1)-2f(x-1)=2x+17
∴kx+5k+b=2x+17
∴k=2
5k+b=17
∴k=2,b=7
∴f(x)=2x+7