Answer
$16$
Work Step by Step
The Mean Value Theorem states that if $f$ is a continuous function over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$, then there exists at least one point $c\in(a,b)$ so that
$$f′(c)=\dfrac{f(b)−f(a)}{b−a}.$$
By the Mean Value Theorem on $[1,4]$ we have:
$f'(c)$ = $\frac{f(4)-f(1)}{4-1}$
$f'(c)$ $\geq$ $2$
$f(4)-f(1)$ = $3f'(c)$
$f(1)$ = $10$
$f(4)$ = $10+3f'(c)$ $\geq$ $10+3(2)$ = $16$
so the smallest possible value of f(4) is $16$.
