1.题不太清楚（f（x-2）2?）,假定为g（x）=3f（x-1）- f（x-2） （x>0)
a.x ≥ 2：x - 1 ≥ 0,x - 2 ≥0,g(x) = 3f(x-1) -f(x-2) = 3*2 - 2 = 4
b.2 > x ≥ 1:x - 1 ≥ 0,x - 2 < 0,g(x) = 3f(x-1) -f(x-2) = 3*2 - 1 = 5
c.0 < x < 1:x - 1 < 0,x - 2 < 0,g(x) = 3f(x-1) -f(x-2) = 3*1 - 1 = 2
d.x ≤ 0:按题,无定义.
2.令t = x+1,x = t -1,f（x+1）=x²-3x+2
f(t) = (t-1)^2 -3(t-1) + 2 = t^2 -5t +6
f(x) = x^2 -5x +6
f(2) = 2^2 -5*2 + 6 = 0
f(a) = a^2 -5a + 6
3.似乎题不全.
f(1) = 1 - m + n = -1,n = m-2
f(x) = x^2 -mx + m-2
f(-5) = 25 -5m +m -2 = 23 -4m
4.f(x) 为一次函数,设为f(x) = mx + n
f(f(x)) = mf(x) + n = m(mx + n) + n = m²x + n(m+1) = 3x + 5
m² = 3,n(m+1) = 5
解为：m = √3,n = 5(1-√3)/2,f(x）= √3x + 5(1-√3)/2
或：m = -√3,n = -5(1+√3)/2,f(x) = -√3x - 5(1+√3)/2
5.题有问题
6.f(x) 为一次函数,设为f(x) = mx + n
f(x+1) = m(x+1) + n = mx + m + n
f(x -1) = m(x-1) + n = mx -m + n
3f(x+1) -2f(x-1) = 3(mx + m + n) - 2(mx -m + n) = mx +5m + n = 2x + 17
m = 2,n = 7
f(x) = 2x + 7
7.函数f（x）是二次函数,设为f(x) = ax² + bx + c (a ≠ 0)
f(0) = c = 0
f(x) = ax² + bx
f(x+1) = a(x+1)² + b(x+1) = ax² + 2ax + a + bx + b = ax² + (2a+b)x + a + b (1)
f(x+1) = f(x) + x+1 = ax² + bx + x + 1 = ax² + (b+1)x + 1 (2)
比较(1)(2)：
2a + b = b + 1
a + b = 1
a = 1/2,b = 1/2
f(x) = (x² + x)/2