Final answer:
The range of possible values for f(12)-f(1) is [121, 143] according to the Mean Value Theorem.
Step-by-step explanation:
To find the range of possible values of f(12)−f(1), we can use the Mean Value Theorem. According to the theorem, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a value c in the open interval (a, b) such that f'(c) is equal to the average rate of change of the function over the interval [a, b]. In this case, the average rate of change of f(x) over the interval [1, 12] is given by (f(12)−f(1))/(12−1) = (f(12)−f(1))/11. Since f'(x) is always between 11 and 13 for all x in the open interval (1, 12), we can conclude that f'(c) is also between 11 and 13. Therefore, 11 ≤ f'(c) ≤ 13. Substituting this into the equation for the average rate of change, we get 11 ≤ (f(12)−f(1))/11 ≤ 13. Multiplying by 11, we have 11² ≤ f(12)−f(1) ≤ 11×13. Simplifying, we get 121 ≤ f(12)−f(1) ≤ 143. Therefore, the range of possible values for f(12)−f(1) is [121, 143].