Now, we are given f (0) = 4 and f'(x) <2 for x > 0. Therefore, for all x 2 0, f(x) < 4 + x (2) = 2x +4 21. Let x > 0. Because f is continuous on [0, x] and differentiable on (0, x), the Mean Value Theorem guarantees there exists a c e (0, x) such that f (x) – f (0) х — 0 f'(c) = f (x) = f (0) + xf'(c) or

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