zmudz one year ago Suppose that $$f(x)$$ is a function such that $$f(xy) + x = xf(y) + f(x)$$ for all real numbers $$x$$ and $$y$$. If $$f(-1) = 5$$ then compute $$f(-1001)$$.

$$f(-1 \times a) = f(a \times -1) = f(-a)$$ x=-1, y = a $$\implies f(-a) -1 = -f(a) + f(-1)$$ $$f(-a) = -f(a) + 6$$ x=a, y = -1 $$\implies f(-a) + a = af(-1) + f(a)$$ $$f(-a) = f(a) + 4a$$ $$f(a) = 3 - 2a$$ $$f(-a) = 3 + 2a$$ 2005 either way