$f\left(x\right)=\{\begin{array}{c}{x}^{2}x\ge 1\\ x^{3}x<1\end{array}$        $g\left(x\right)=\{\begin{array}{cc}2x-3x>0& \\ x-4x\le 0& \end{array}$

Determinare l'espressione analitica di $f\circ g$ e $g\circ f$.

$f\circ g=f\left(g\left(x\right)\right)=\{\begin{array}{c}{\left[g\left(x\right)\right]}^{2}g\left(x\right)\ge 1\\ {\left[g\left(x\right)\right]}^{3}g\left(x\right)<1\end{array}$

$g\left(x\right)\ge 1\Rightarrow$$\{\begin{array}{c}2x-3\ge 1x>0\\ x-4\ge 1x\le 0\end{array}\Rightarrow \{\begin{array}{c}x\ge 2x>0\\ x\ge 5x\le 0\end{array}\Rightarrow \mathrm{x\ge }2$

$f\left(g\left(x\right)\right)=\{\begin{array}{c}{\left[g\left(x\right)\right]}^{2}x\ge 2\\ {\left[g\left(x\right)\right]}^{3}x<2\end{array}=\{\begin{array}{c}{\left(2x-3\right)}^{2}x\ge 2\\ \{\begin{array}{c}{\left(2x-3\right)}^{3}0<x<2\\ {\left(x-4\right)}^{3}x\le 0\end{array}\end{array}$

$g\circ f=g\left(f\left(x\right)\right)=$$\{\begin{array}{c}2f\left(x\right)-3f\left(x\right)>0\\ f\left(x\right)-4f\left(x\right)\le 0\end{array}$

$f\left(x\right)>0\Rightarrow$$\{\begin{array}{c}{x}^2>0x\ge 1\\ x^3>0x<1\end{array}\Rightarrow \{\begin{array}{c}x\ne 0x\ge 1\\ x>0x<1\end{array}\Rightarrow \{\begin{array}{c}x\ge 1\\ 0<x<1\end{array}\Rightarrow x>0$

$g\left(f\left(x\right)\right)=\{\begin{array}{c}2f\left(x\right)-3x>0\\ f\left(x\right)-4x\le 0\end{array}=\begin{array}{c}\{\begin{array}{c}\{\begin{array}{c}2x^2-3x\ge 1\\ 2x^3-30<x<1\end{array}\\ x^3-4x\le 0\end{array}\end{array}$