令 t= x-1 则 x=1+t
原式可为
3f(t) + 2f(-t)=2+2*t .(1)
将-t放入等式
3f(-t) + 2f(t) = 2 + 2 * (-t).(2)
(1)+(2) 得
5f(t) + 5f(-t) = 4 ---> f(t) + f(-t) = 4/5 -------(3)
(1)-(2) 得
f(t) - f(t) = 4t .(4)
(3) + (4) 得
2f(t)=4t + 4/5
所以
f(t)= 2t + 2/5
即
f(x)=2x+ 2/5
令 t= x-1 则 x=1+t
原式可为
3f(t) + 2f(-t)=2+2*t .(1)
将-t放入等式
3f(-t) + 2f(t) = 2 + 2 * (-t).(2)
(1)+(2) 得
5f(t) + 5f(-t) = 4 ---> f(t) + f(-t) = 4/5 -------(3)
(1)-(2) 得
f(t) - f(t) = 4t .(4)
(3) + (4) 得
2f(t)=4t + 4/5
所以
f(t)= 2t + 2/5
即
f(x)=2x+ 2/5