Question

已知函数f(x)是一次函数,且2(f)+3f(2)=3,2f(-1)-f(0)=-1,求f(x)的解析式.

Answer 1

设f(x)=kx+b

则f(1)=k*1+b=k+b

f(2)=k*2+b=2k+b

f(-1)=k*(-1)+b=-k+b

f(0)=k*0+b=b

2f(1)+3f(2)=3,2f(-1)-f(0)=-1

所以2(k+b)+3(2k+b)=3

8k+5b=3 (1)

2(-k+b)-b=-1

-2k+b=-1 (2)

(1)+(2)*4

8k+5b-8k+4b=3+(-1)*4

9b=-1

b=-1/9

k=(b+1)/2=4/9

f(x)=(4/9)x-1/9