The minimum is obviously obtained for some negative $x{}$. So we only consider this domain. Using $f(x)=\tfrac{x\cdot f(x)-f(x)}{x-1}$, we get \[f(x):=x^{2014} + 2x^{2013} +\cdots + 2014x + 2015=\frac{x^{2016}-2016 x+2015}{(x-1)^2},\]and so \[f^\prime(x)=\frac{2014 x^{2016}-2016 x^{2015}+2016 x-2014}{(x-1)^3}.\]For $x<-1$, we have $2014x^{2016}-2014>0$ and $2016x-2016x^{2015}>0$, $(x-1)^3<0$, so $f^\prime(x)<0$. For $x=-1$, $f^\prime(x)=0$. For $x\in(-1,0)$, we have $2014x^{2016}-2014<0$ and $2016x-2016x^{2015}<0$, $(x-1)^3<0$, so $f^\prime(x)>0$. Hence $x=-1$ yields minimum $f(-1)=1008$.