As $x$ increases, $(2)^x$ increases, so $g(x)$ increases. The minimum value occurs as $x \to -\infty$, where $(2)^x \to 0$. Thus $g(x) \to \dfrac{1}{11}(6 \cdot 0 + 7) = \dfrac{7}{11}$. Since $g(x) > k$ for all $x$, the greatest $k$ is $\dfrac{7}{11}$.