If the domain of the function  $f\left(x\right)={\log }_e(4x^2+11x+6)+{\sin }^{-1}(4x+3)+{\cos }^{-1}\left(\frac{10x+6}{3}\right)$  is  $(\alpha ,\beta ]$ . Then  $8|\alpha +\beta |$  is equal to

$f\left(x\right)={\log }_e(4x^2+11x+6)+{\sin }^{-1}(4x+3)+{\cos }^{-1}\left(\frac{10x+6}{3}\right)$
${\log }_e(4x^2+11x+6)$
$\begin{array}{l}\text{⇒}4x^2+11x+6>0\\ 4x^2+8x+3x+6>0\\ (4x+3)(x+2)>0\\ x\in (-\mathrm{\infty },-2)\cup \left(-\frac{3}{4},\mathrm{\infty }\right)\end{array}$
${\sin }^{-1}(4x+3)$
$$\begin{gathered} \begin{array}{l}\text{⇒ -1≤}4x+3\le 1 \\ \Rightarrow -4\le 4x\le 2\\ \Rightarrow x\in [-1,-1/2]\end{array} \end{gathered}$$
${\cos }^{-1}\left(\frac{10x+6}{3}\right)$
$$\begin{gathered} \begin{array}{l}\text{⇒-1≤}\frac{10x+6}{3}\text{≤1} \\ \Rightarrow -9\le 10x\le -3\\ \Rightarrow x\in \left[-\frac{9}{10},-\frac{3}{10}\right]\\ \Rightarrow x\in \left(-\frac{3}{4},-\frac{1}{2}\right]\\ \alpha =-\frac{3}{4},\beta =-\frac{1}{2}\\ \alpha +\beta =-\frac{5}{4}\end{array} \end{gathered}$$