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

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

$f\circ g=f\left(g\left(x\right)\right)=\{\begin{array}{c}{\left[g\left(x\right)\right]}^{2}g\left(x\right)\le 1\\ \sqrt{g\left(x\right)}g\left(x\right)>1\end{array}$

$g\left(x\right)\le 1\Rightarrow$$\{\begin{array}{c}x+1\le 1x<0\\ 3x\le 1x\ge 0\end{array}\Rightarrow \{\begin{array}{c}x\le 0x<0\\ x\le \frac{1}{3}x\ge 0\end{array}\Rightarrow x\le \frac{1}{3}$

$f\left(g\left(x\right)\right)=\{\begin{array}{c}{\left[g\left(x\right)\right]}^{2}x\le \frac{1}{3}\\ \sqrt{g\left(x\right)}x>\frac{1}{3}\end{array}=\{\begin{array}{c}\{\begin{array}{c}{\left(x+1\right)}^{2}x<0\\ {9x}^{2}0\le x\le \frac{1}{3}\end{array}\\ \sqrt{3x}x>\frac{1}{3}\end{array}\begin{array}{c}\end{array}$

$g\circ f=g\left(f\left(x\right)\right)=$$\{\begin{array}{c}f\left(x\right)+1f\left(x\right)<0\\ 3f\left(x\right)f\left(x\right)\ge 0\end{array}$

$f\left(x\right)<0\Rightarrow$$\{\begin{array}{c}{x}^2<0x\le 1\\ \sqrt{x}<0x>1\end{array}\Rightarrow \{\begin{array}{c}\nexists x\in ℝx\le 1\\ \nexists x\in ℝx>1\end{array}\Rightarrow \nexists x\in ℝ$

$g\left(f\left(x\right)\right)=\{\begin{array}{c}f\left(x\right)+1\nexists x\in ℝ\\ 2f\left(x\right)\forall x\in ℝ\end{array}=\{\begin{array}{c}2x^{2}x<1\\ 2\sqrt{x}x>1\end{array}\begin{array}{}\end{array}$