Final answer:
Using the Mean Value Theorem, we can prove that if a function f is differentiable and non-decreasing (f'(x) >= 0), with at least one point of positive derivative, then f(b) will indeed be greater than f(a) for any b > a.
Step-by-step explanation:
To prove that if a function f is differentiable, has a derivative greater than or equal to 0 across its domain, and has at least one point where the derivative is positive, then f(b) > f(a) for any b > a, we invoke the Mean Value Theorem (MVT). The MVT states that if f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists some c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
Since the derivative f'(x) >= 0 for all x, the function f is non-decreasing. Furthermore, because there is at least one point where the derivative is positive, the function f is strictly increasing at that point, which implies that there must be an interval where f is strictly increasing, therefore f(b) - f(a) > 0 implying f(b) > f(a).
By the conditions given, since there is a point with positive derivative, say at x = c, then according to the MVT there is an interval where f' implies growth, and f(b) > f(a). Hence, the original statement is proven.