3f(x)-5f(x-1)=x. find f(x)+f(1-x)
Since the given difference of two constant multiples of f(x) is a polynomial, then so also is f(x) (and f(x-1) ) a polynomial.
Since the polynomial on the rhs is a linear function, then let f(x) be a general linear function.
i.e. let f(x) = ax + b
Then, f(x-1) = a(x-1) + b
Substituting for f(x) and f(x-1) into the original expression,
3f(x) - 5f(x-1) = x
3(ax + b) – 5(a(x-1) + b) = x
3ax + 3b – 5ax + 5a – 5b = x
-2ax + 5a – 2b = x
Comparing coefficients,
-2a = 1
5a – 2b = 0
Giving a = -1/2, b = -5/4
This gives f(x) = -x/2 – 5/4
Then f(1 – x) = -(1 – x)/2 – 5/4 = -1/2 + x/2 – 5/4 = x/2 – 7/4
So, f(x) + f(1 – x) = -x/2 – 5/4 + x/2 – 7/4 = -12/4 = -3
Answer: f(x) + f(1 – x) = -3