show that function g is constant and f is f(x)=mx+b if f(x)=(x-y)g(x)+f(y)?

You need to prove that the function `g(x)` is a constant,
considering the relation provided by the problem, such
that:

`f(x) = (x-y)g(x) + f(y) => f(x) - f(y) = (x -
y)g(x)`

`g(x) = (f(x) - f(y))/(x - y)`

You need to remember the mean value theorem such
that:

`f'(c) = (f(x) - f(y))/(x - y)` , if `c in [x,y]`

Since `f'(c) = (f(x) - f(y))/(x - y)` , hence `f'(c) =
g(x)` , thus, you need to evaluate the derivative of the function f(x), such
that:

`f'(x) = (mx + b)' => f'(x) = m`

Notice that the derivative of the function `f(x)` is a
constant m, hence `f'(c) = g(x) = m.`

Hence,
checking if the function g(x) is a constant, using the mean value theorem, yields that
g`(x) = f'(x) = m =` constant.