Question

(1)已知f(x)是二次函数，若f(0)=0,且f(x+1)=f(x)+x+1,求f(x)的解析式。

(1)已知f(x)是二次函数，若f(0)=0,且f(x+1)=f(x)+x+1,求f(x)的解析式。
(2)已知3f(x)+5f(1/x)=2/x+1,求函数f(x)的解析式。
要详细过程谢谢

Answer 1

(1)
f(0) = 0, f(x) = ax² + bx
f(x+ 1) = a(x + 1)² + b(x + 1) = ax² + (2a + b)x + a + b = f(x) + x + 1 = ax² + (b + 1)x + 1
比较系数:
2a + b = b + 1, a = 1/2
a + b = 1, b = 1 - a = 1/2
f(x) = (x² + x)/2

(2)
3f(x) + 5f(1/x) = 2/x + 1 (i)
用1/x取代x:
3f(1/x) + 5f(x) = 2/(1/x) + 1 = 2x + 1 (ii)
5*(ii) - 3*(i):
16f(x) = 10x - 6/x + 2
f(x) = 5x/8 - 3/(8x) + 1/8