EXAMPLE 5 Suppose that f(0) = −4 and f '(x) ≤ 5 for all values of x. How large can f(2) 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, 2] . There exists a number c such that f(2) − f(0) = f '(c) − 0 so f(2) = f(0) + f '(c) = −4 + f '(c). We are given that f '(x) ≤ 5 for all x, so in particular we know that f '(c) ≤ . Multiplying both sides of this inequality by 2, we have 2f '(c) ≤ , so f(2) = −4 + f '(c) ≤ −4 + = . The largest possible value for f(2) is .