যদি   $f{\left ( x \right )} = x^{2} , g{\left ( x \right )} = x^{3} + 1 , h \left ( x \right ) = x + 2 , x = - 3$  হয় তবে hogof নির্ণয় করো এবং দেখাও যে gof≠fog.

1st Part :

$\begin{array}{l} \text { hogof }=\mathrm{h}[\mathrm{g}\{\mathrm{f}(\mathrm{x})\}] \\ \qquad \begin{array}{l} \quad=\mathrm{h}\left[\mathrm{g}\left\{\mathrm{x}^{2}\right\}\right] \\ =\mathrm{h}\left[\left(\mathrm{x}^{2}\right)^{3}+1\right]=\mathrm{h}\left[\mathrm{x}^{6}+1\right] \\ \quad=\mathrm{x}^{6}+1+2=\mathrm{x}^{6}+3 \end{array} \\ \therefore \text { hogof }(-3)=(-3)^{6}+3 \quad[\because \mathrm{x}=-3] \\ =729+3=732 \quad \text { [Ans. }] \end{array}$

2nd Part :

$\begin{array}{l} \text { gof }=g\{f(x)\} \\ =g\left\{x^{2}\right\}=\left(x^{2}\right)^{3}+1=x^{6}+1 \\ f o g=f\{g(x)\}=f\left\{x^{3}+1\right\} \\ =\left(x^{3}+1\right)^{2}=x^{6}+2 x^{3}+1 \\ \therefore g o f \neq \text { fog [Showed] } \end{array}$