Question

If the f(x)=|X| + X and ,g(x)= |X | - X ,find fog (-3) and gof(-2 ).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:f(x) = |x| + x}$

and

$\red{\rm :\longmapsto\:g(x) = |x| - x}$

Consider,

$\rm :\longmapsto\:fog( - 3)$

$\rm \: = \: f\bigg[g( - 3)\bigg]$

$\rm \: = \: f\bigg[ | - 3| - ( -3 ) \bigg]$

$\rm \: = \: f\bigg[3 + 3\bigg]$

$\rm \: = \: f(6)$

$\rm \: = \: |6| + 6$

$\rm \: = \: 6 + 6$

$\rm \: = \: 12$

$\bf\implies \: \red{\boxed{ \bf \:\:fog( - 3) = 12}}$

Now, Consider,

$\red{\rm :\longmapsto\:gof( - 2)}$

$\rm \: = \: g\bigg[f( - 2)\bigg]$

$\rm \: = \: g\bigg[ | - 2| - 2\bigg]$

$\rm \: = \: g\bigg[2 - 2\bigg]$

$\rm \: = \: g\bigg[0\bigg]$

$\rm \: = \: |0| - 0$

$\rm \: = \: 0$

$\bf\implies \: \red{\boxed{ \bf \:\:gof( - 2) = 0}}$

$\purple{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{fog( - 3) = 12} \\ \\ &\sf{gof( - 2) = 0} \end{cases}\end{gathered}\end{gathered}}$