If \[f\left( x \right) = \left\{ \begin{array}{l} {x^3} + 1,\,\,\,x < 0\\ {x^2} + 1,\,\,\,x \ge 0 \end{array} \right.,\,g\left( x \right) = \left\{ \begin{array}{l} {\left( {x - 1} \right)^{\frac{1}{3}}},\,\,\,x < 1\\ {\left( {x - 1} \right)^{\frac{1}{2}}},\,\,\,x \ge 1 \end{array} \right.,\]           then $$\left( {gof} \right)\left( x \right)$$   is equal to :

Solution :
$$\eqalign{ & {\text{Let }}x < 0 \cr & \therefore \,\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) \cr & = g\left( {{x^3} + 1} \right) \cr & = {\left[ {\left( {{x^3} + 1} \right) - 1} \right]^{\frac{1}{3}}}\,\,\left( {\because x < 0 \Rightarrow {x^3} + 1 < 1} \right) \cr & = {\left( {{x^3}} \right)^{\frac{1}{3}}} \cr & = x \cr & {\text{Let }}x \geqslant 0 \cr & \therefore \,\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) \cr & = g\left( {{x^2} + 1} \right) \cr & = {\left( {\left( {{x^2} + 1} \right) - 1} \right)^{\frac{1}{2}}}\,\,\left( {\because x \geqslant 0 \Rightarrow {x^2} + 1 \geqslant 1} \right) \cr & = {\left( {{x^2}} \right)^{\frac{1}{2}}} \cr & = \left| x \right| \cr & = x\,\,\,\,\,\,\left( {\because x \geqslant 0} \right) \cr & \therefore \,\left( {gof} \right)\left( x \right) = x,\,\forall \,x\, \in \,R \cr} $$