Solution: To solve this, we need to set the function greater than or equal to zero and use our standard inequality techniques to solve:

\[ \solve{ -11\left(x-\dfrac{{1}}{{7}}\right)\left(x-\dfrac{{6}}{{11}}\right)&\gt&0 } \] From here we solve the sub-problem identify the actual zeros: \(x=\frac{{1}}{{7}}\) and \(x=\frac{{6}}{{11}}\). Then we construct our table to determine all the sign changes. Don't forget the original negative!! \[ \begin{{array}}{ |c|c|c|c| } \hline &\left(-\infty, \dfrac{{1}}{{7}}\right)&\left(\dfrac{{1}}{{7}},\dfrac{{6}}{{11}}\right)&\left(\dfrac{{6}}{{11}},\infty\right) \\ \hline -11&\mathbb{-}&\mathbb{-}&\mathbb{-}\\ \hline x-\dfrac{{1}}{{7}}&\mathbb{-}&\mathbb{+}&\mathbb{+}\\ \hline x-\dfrac{{6}}{{11}}&\mathbb{-}&\mathbb{-}&\mathbb{+}\\ \hline &\downarrow&\downarrow&\downarrow \\ \hline \text{Result:}&\mathbb{-}&\mathbb{+}&\mathbb{-}\\ \hline &{\color{{red}} \times}&{\color{{green}} \checkmark}&{\color{{red}} \times} \\ \hline \end{{array}} \] Thus, our answer is \( \left(\dfrac{{1}}{{7}},\dfrac{{6}}{{11}}\right)\). Once we know our actual values, it is easy to verify by looking at the graph on the window \(x=-1\) to \(x=1\) and \(y=-1\) to \(y=1\).