Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (k)/Solution 1

Notice that $\displaystyle f(x)$ is differentiable on $\displaystyle [0,4]$ and hence it is continuous there. Thus, by the mean value theorem, there exists a point $\displaystyle c\in [0,4]$ such that

$\displaystyle f'(c)={\frac {f(4)-f(0)}{4-0}}$.

We know that on $\displaystyle [0,4]$ that $\displaystyle f'(x)\geq 3$. Since $\displaystyle c\in [0,4]$ and $\displaystyle f(0)=10$ , we have that

$\displaystyle 3\leq f'(c)={\frac {f(4)-f(0)}{4-0}}={\frac {f(4)-10}{4}}$.

Solving gives $\displaystyle 22\leq f(4)$ and hence the minimum $\displaystyle f(4)$ could be is 22.