Video Example EXAMPLE 5 Suppose that f(0) = -3 and f'(x) < 5 for all values of x. How large can f(3) possibly be? SOLUTION We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval [0, 3]. There exists a number c such that - f(3) − f(0) = f'(c)( - 0) so f(3) = f(0) + f'(c)=-3+ f'(c). We are given that f'(x) 5 for all x, so in particular we know that f'(c) < of this inequality by 3, we have 3f'(c) < SO f(3) = -3+ f'(c) ≤ −3+ The largest possible value for f(3) is = Multiplying both sides.