\(B=1\dfrac{2}{3}-\left|\dfrac{1}{3}x+4\right|=\dfrac{5}{3}-\left|\dfrac{1}{3}x+4\right|\)
\(=\left[{}\begin{matrix}\dfrac{5}{3}-\dfrac{1}{3}x-4=-\dfrac{1}{3}x-\dfrac{7}{3}\left(x\ge-12\right)\\\dfrac{5}{3}+\dfrac{1}{3}x+4=\dfrac{1}{3}x+\dfrac{17}{3}\left(x< -12\right)\end{matrix}\right.\)
