When we open the brackets of an absolute value we get the following:
[tex]\begin{gathered} \lvert g(x)\rvert \\ g(x),g(x)\ge0\text{ and }-g(x),g(x)<0 \end{gathered}[/tex]So in this case we have:
[tex]\lvert11-3x\rvert[/tex]After open the absolute balue we get two expressions. The first one is:
[tex]11-3x,\text{ }11-3x>0[/tex]And the second one is:
[tex]-11+3x,\text{ }11-3x\leq0[/tex]So let's work with the inequalities of each case. The one in the first case is:
[tex]11-3x>0[/tex]We can add 3x to both sides:
[tex]\begin{gathered} 11-3x+3x>0+3x \\ 11>3x \end{gathered}[/tex]And we divide both sides by 3:
[tex]\begin{gathered} \frac{11}{3}>\frac{3x}{3} \\ \frac{11}{3}>x \end{gathered}[/tex]The inequality in the second case is:
[tex]11-3x\leq0[/tex]We can add 3x to both sides:
[tex]\begin{gathered} 11-3x+3x\leq0+3x \\ 11\leq3x \end{gathered}[/tex]And divide by 3:
[tex]\begin{gathered} \frac{11}{3}\leq\frac{3x}{3} \\ \frac{11}{3}\leq x \end{gathered}[/tex]Then the two parts of the function are:
[tex]\begin{gathered} 11-3x,x<\frac{11}{3} \\ -11+3x,x\ge\frac{11}{3} \end{gathered}[/tex]Then the answers are the second and fourth options.