Mark M. answered 05/20/21
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Since 5 ≤ f'(x) ≤ 9 for all x, f(x) is differentaible and continuous for all x (differentiability implies continuity)
So, f(x) is continuous on [3,12] and is differentiable on (3,12). Thus, by the Mean Value Theorem, there is at least one number c in (3,12) such that f'(c) = [f(12) - f(3)] / (12 - 3).
So, f(12) - f(3) = 9f'(c)
But, 5 ≤ f'(c) ≤ 9
So, 45 ≤ 9f'(c) ≤ 81
Therefore, 45 ≤ f(12) - f(3) ≤ 81