If f, g : R → R be two functions defined as f(x) = |x| + x and g(x)

Solution:

Given: $f(x)=|x|+x$

and $g(x)=|x|-x, \forall x \in \mathrm{R}$

$f o g=f(g(x))=|g(x)|+g(x)$

$=|| x|-x|+(|x|-x)$

Therefore,

$f(g(x))= \begin{cases}0 & x \geq 0 \\ 4 x & x<0\end{cases}$

$f o g= \begin{cases}4 x & x<0 \\ 0 & x \geq 0\end{cases}$

$g o f=g(f(x))=|f(x)|-f(x)$

$=|| x|+x|-(|x|+x)$

$g(f(x))= \begin{cases}0 & x \geq 0 \\ 0 & x<0\end{cases}$

Therefore, $g(f(x))=g o f=0$

Now, $f o  g(-3)=(4)(-3)=-12 \quad$ (since, $f \circ g=4 x$ for $x<0)$

$f o g(5)=0 \quad$ (since, $f o g=0$ for $x \geq 0)$

$g o  f(-2)=0 \quad$ (since, $g \circ f=0$ for $x<0)$