If the domain of the function
$\mathrm{f}\left(\mathrm{x}\right)={\log }_{\mathrm{e}}\left(4{\mathrm{x}}^2+11\mathrm{x}+6\right)+{\sin }^{-1}(4\mathrm{x}+3)+{\cos }^{-1}\left(\frac{10x+6}{3}\right)\text{ is }(\mathrm{\alpha },\mathrm{\beta }]\text{ then }36|\mathrm{\alpha }+\mathrm{\beta }|=$

$\mathrm{f}\left(\mathrm{x}\right)={\log }_{\mathrm{e}}\left(4{\mathrm{x}}^2+11\mathrm{x}+6\right)+{\sin }^{-1}(4\mathrm{x}+3)+{\cos }^{-1}\left(\frac{10\mathrm{x}+6}{3}\right)$$$\begin{gathered} \begin{array}{l}{\log }_{\mathrm{e}}\left(4{\mathrm{x}}^2+11\mathrm{x}+6\right) is definied if \\ 4{\mathrm{x}}^2+11\mathrm{x}+6>0 \\ \Rightarrow (\mathrm{x}+2)(4\mathrm{x}+3)>0 \\ \Rightarrow \mathrm{x}<-2\text{ or }\mathrm{x}>\frac{-3}{4}\to \left(1\right)\\ and {\sin }^{-1}(4\mathrm{x}+3) is defined if \\ -1\le 4\mathrm{x}+3\le 1 \\ \Rightarrow -1\le \mathrm{x}\le -\frac{1}{2}\to \left(2\right) \\ {\cos }^{-1}\left(\frac{10\mathrm{x}+6}{3}\right) is defined if -1\le \frac{10\mathrm{x}+6}{3}\le 1\\ \Rightarrow \frac{-9}{10}\le \mathrm{x}\le \frac{-3}{10}\to \left(3\right) \\ \text{from }\left(1\right)\left(2\right)\left(3\right),\mathrm{x}\in \left[\frac{-3}{4},\frac{-1}{2}\right] \\ 36|\mathrm{\alpha }+\mathrm{\beta }|=36\left|\frac{-3}{4}-\frac{1}{2}\right|=45\end{array} \end{gathered}$$