# What are functions f(x+y)+2f(c-y)=3f(x)-y

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You should come up with the substitution y = x such that:

f(x+x) + 2f(x-x) = 3f(x) - x

=> f(2x) + 2f(0) = 3f(x) - x

You should come up with the substitution y = -x such that:

f(x-x) + 2f(x + x) = 3f(x) + x

=> f(0) + 2f(2x) = 3f(x) + x

You need to eliminate f(2x).

Multiply the first relation you have obtained by 2 such that:

2f(2x) + 4f(0) = 6f(x) - 2x

Subtracting the relation f(0) + 2f(2x) = 3f(x) + x from 2f(2x) + 4f(0) = 6f(x) - 2x yields:

2f(2x) + 4f(0) - f(0) - 2f(2x) = 6f(x) - 2x - 3f(x) - x

3f(0) = 3f(x) - 3x => f(0) = f(x) - x

=> f(x) = f(0) + x

**The function f(0)=a denotes a constant, hence the functions that check the relation f(x+y)+2f(x-y)=3f(x)-y are: f(x) = x + a.**