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

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

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

$g\left(x\right)\ge 1\Rightarrow$$\{\begin{array}{c}x+1\ge 1x\ge 1\\ x^2\ge 1x<1\end{array}\Rightarrow \{\begin{array}{c}x\ge 0x\ge 1\\ x\le -1\vee x\ge 1x<1\end{array}\Rightarrow \{\begin{array}{c}x\ge 1\\ x\le -1\end{array}\Rightarrow x\le -1\vee x\ge 1$

$f\left(g\left(x\right)\right)=\{\begin{array}{c}\sqrt{g\left(x\right)}x\le -1\vee x\ge 1\\ 2g\left(x\right)-1<x<1\end{array}=\{\begin{array}{c}\{\begin{array}{c}\sqrt{x+1}x\ge 1\\ \sqrt{x^2}x\le -1\end{array}\\ 2x^2-1<x<1\end{array}\begin{array}{c}\end{array}$

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

$f\left(x\right)\ge 1\Rightarrow$$\{\begin{array}{c}\sqrt{x}\ge 1x\ge 1\\ 2x\ge 1x<1\end{array}\Rightarrow \{\begin{array}{c}x\ge 1x\ge 1\\ x\ge \frac{1}{2}x<1\end{array}\Rightarrow \{\begin{array}{c}x\ge 1\\ \frac{1}{2}\le x<1\end{array}\Rightarrow x\ge \frac{1}{2}$

$g\left(f\left(x\right)\right)=\{\begin{array}{c}f\left(x\right)+1x\ge \frac{1}{2}\\ f^2\left(x\right)x<\frac{1}{2}\end{array}=\begin{array}{c}\{\begin{array}{c}\{\begin{array}{c}\sqrt{x}+1x\ge 1\\ 2x+1\frac{1}{2}\le x<1\end{array}\\ 4x^{2}x\le \frac{1}{2}\end{array}\end{array}$