Minimum Value

Note that, $-\sqrt{a^2+b^2}\leq a\sin\theta+b\cos\theta\leq \sqrt{a^2+b^2}$ Hence minimum value of the function is $-5+2014=\boxed{2009}$

Note first that

$3\sin(x) + 4\cos(x) = 5*((\frac{3}{5})\sin(x) + (\frac{4}{5})\cos(x)) = 5*\sin(x + \arccos(\frac{3}{5}))$ .

Thus, since the minimum of $\sin(\theta)$ is $-1$ , the minimum of $3\sin(x) + 4\cos(x)$ is $5*(-1) = -5$ , and thus the minimum of $f(x)$ is $(-5) + 2014 = \boxed{2009}$ .

(Alternately, we have that $f'(x) = 0$ when $\tan(x) = \frac{3}{4}$ , which leads to minimum co-values for $\sin(x)$ and $\cos(x)$ of $-\frac{3}{5}$ and $-\frac{4}{5}$ , respectively. This results in the same minimum value for $f(x)$ as found above.)