Final answer:
Using the Mean Value Theorem and the fact that f"(x) ≥2, we deduced that the smallest slope f'(c) at x=2 is 2. Integrating this minimum slope over the interval [2, 5], we found that the smallest possible value for f(5) is 12.
Step-by-step explanation:
The question is concerned with finding the minimum possible value of f(5) given that f(2) = 6 and f"(x) ≥2 for the interval 2≤x≤5. By the Mean Value Theorem, there exists a point c in (2, 5) such that f'(c) = (f(5) - f(2)) / (5 - 2). Since f"(x) ≥2, this implies that f'(x) is increasing on the interval [2, 5].
Therefore, the smallest possible slope f'(c) can have at x = 2 is 2, the minimum of f"(x). Integrate from 2 to 5, the smallest f'(x) over this interval gives us f'(x) = 2x+C, where C is the integration constant. We can solve for C using the fact that f(2) = 6.
Since f'(x) = 2x + C and f(2) = 6, we integrate f'(x) to get f(x) = x^2 + Cx + D. Using f(2) = 6, we find C and D. Finally, we calculate f(5) which will be 5^2 + 5C + D. Since we want the smallest f(5), we use the smallest possible f'(x), which is a constant function f'(x) = 2, because f"(x) ≥2. Integrating f'(x) = 2 from x = 2 to x = 5 gives us f(5) = f(2) + 3*2 = 6 + 6 = 12. So the smallest value f(5) can be is 12.