Answer:
Given, \[f(x)=|x|+x,\] which can be redefined as \[f(x)=\left\{ \begin{align} & 2x,\,\,\text{if}\,\,x\ge 0 \\ & 0,\,\,\text{if}\,\,x<0 \\ \end{align} \right.\] and \[g(x)=\,\,|x|-x\] \[\Rightarrow \] \[g(x)=\left\{ \begin{align} & 0,\,\,\text{if}\,\,x\ge 0 \\ & -2x,\,\,\text{if}\,\,x<0 \\ \end{align} \right.\] Now, gof gets defined as For \[x\ge 0,\] \[(gof)\,(x)=g(f(x))=g(2x)=0\] and for \[x<0,\] \[(gof)\,(x)=g(f(x))=g(0)=0\] Consequently, we have (gof)(x) = 0, \[\forall x\in R\] Similarly, fog gets defined as For \[x\ge 0,\] (fog)(x) = f(g(x)) = f(0) = 0 and for \[x\ge 0,\] \[(fog)(x)=f(g(x))=f(-2x)=-\,4x\] i.e. \[(fog)\,(x)=\left\{ \begin{matrix} 0, & x\ge 0 \\ -\,4x, & x<{{0}^{.}} \\ \end{matrix} \right.\]
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