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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 \(f(-1001)\).
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\(f(-1 \times a) = f(a \times -1) = f(-a) \) 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
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