N° 4 PROPRIETÀ FUNZIONI

Date le funzioni g ed f così definite:

$f (x) = {\begin{matrix} 2 x > 0 \\ 3 x + 2 x \leq 0 \end{matrix}$ $g (x) = {\begin{matrix} \sqrt{x - 1} x \geq 1 \\ x - 1 x < 1 \end{matrix}$

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

$f \circ g = f (g (x)) = {\begin{matrix} 2 g (x) > 0 \\ 3 g (x) + 2 g (x) \leq 0 \end{matrix}$

$g (x) > 0 \Rightarrow$ ${\begin{matrix} \sqrt{x - 1} > 0 x \geq 1 \\ x - 1 > 0 x < 1 \end{matrix} \Rightarrow {\begin{matrix} x > 1 x \geq 1 \\ x > 1 x < 1 \end{matrix} \Rightarrow x > 1$

$f (g (x)) = {\begin{matrix} 2 x > 1 \\ {\begin{matrix} 3 \sqrt{x - 1} + 2 x = 1 \\ 3 (x - 1) + 2 x < 1 \end{matrix} \end{matrix} \begin{matrix} \end{matrix} = \begin{matrix} \end{matrix} {\begin{matrix} 2 x > 1 \\ {\begin{matrix} 2 x = 1 \\ 3 x - 1 x < 1 \end{matrix} \end{matrix} = {\begin{matrix} 2 x \geq 1 \\ 3 x - 1 x < 1 \end{matrix}$

$g \circ f = g (f (x)) =$ ${\begin{matrix} \sqrt{f (x) - 1} f (x) \geq 1 \\ f (x) - 1 f (x) < 0 \end{matrix}$

$f (x) \geq 1 \Rightarrow$ ${\begin{matrix} 2 \geq 1 x > 0 \\ 3 x + 2 \geq 1 x \leq 0 \end{matrix} \Rightarrow {\begin{matrix} \forall x \in ℝ x > 0 \\ x \geq - \frac{1}{3} x \leq 0 \end{matrix} \Rightarrow x \geq - \frac{1}{3}$

$g (f (x)) = {\begin{matrix} \sqrt{f (x) - 1} x \geq - \frac{1}{3} \\ f (x) - 1 x < - \frac{1}{3} \end{matrix} = {\begin{matrix} {\begin{matrix} 1 x > 0 \\ \sqrt{3 x + 2 - 1} = \sqrt{3 x + 1} - \frac{1}{3} \leq x \leq 0 \end{matrix} \\ 3 x + 2 - 1 = 3 x + 1 x < - \frac{1}{3} \end{matrix} \begin{matrix} \end{matrix}$