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

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

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

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