Science talk:Math Exam Resources/Courses/MATH100/December 2011/Question 1 (l)

Alternatively, you can use the MVT (Mean Value Theorem), which states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

f '(c) = (f(b)-f(a))/(x-a)

In this case, the formula we use will be:

(f(2)-f(-1))/(2-(-1)) = f'(c)

Plugging in known values and simplifying gives: (f(2)-9)/3 >= 3

Solving for f(2) we get f(2) >= 18

Therefore the minimal value of f(2) is 18.

I'll add this - it's a nice elegant solution!