Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (l)/Solution 2

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

$f'(c)={\frac {f(b)-f(a)}{b-a}}$

If we assume $f(x)$ is differentiable on (-1,2) and differentiable and continuous on the interval [-1,2], we can use the MVT which gives:

${\frac {f(2)-f(-1)}{2-(-1)}}=f'(c)$

For some value c in the interval (-1,2). However, we know that $f'(x)\geq 3$ for any value between -1 and 2, thus, $f'(c)\geq 3$. Adding this, plugging in known values, and simplifying gives:

${\frac {f(2)-9}{3}}\geq 3$

Solving for f(2) we get $f(2)\geq 18$, thus the minimal value of f(2) is 18.