By induction f(mx) = m(f(x)+C)-C.

Let x=1/n, m=n and find that f(1/n) = (1/n)(f(1)+C)-C.

Now let x=1/n and find that f(m/n) =(m/n)(f(1)+C)-C. f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x)(since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C.

Since f is monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem).

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