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ciao ho delle difficoltà con lo studio della seguente funzione:

\[f(x) = x\,\sqrt {\frac{{2x}}{{x + 3}}} \]

se mi può aiutare.
la ringrazio.

Ciao, \[f(x) = x\,\sqrt {\frac{{2x}}{{x + 3}}} \]
ecco i passi fondamentali per lo studio della funzione tralasciando i dettagli:

- Studio del dominio e analisi punti singolari
\[C.E.\left\{ {\left. \begin{array}{l}
\frac{{2x}}{{x + 3}} \ge 0\\
x + 3 \ne 0
\end{array} \right\}} \right.\;\; \to \;\;{D_f} = ] - \infty , - 3\,[\, \cup \,\,[0, + \infty \,[\]
\[{x_0} = - 3\;\;\left\{ \begin{array}{l}
f( - 3) = \not \exists \\
\mathop {\lim }\limits_{x \to - {3^ - }} x\,\sqrt {\frac{{2x}}{{x + 3}}} = - 3\,\sqrt {\frac{{ - 6}}{{{0^ - }}}} = - 3\sqrt { + \infty } = - \infty \;\;(A.V.)\\
\mathop {\lim }\limits_{x \to - {3^ + }} x\,\sqrt {\frac{{2x}}{{x + 3}}} = \not \exists
\end{array} \right.\]
\[{x_0} = 0\;\;\left\{ \begin{array}{l}
f(0) = 0\\
\mathop {\lim }\limits_{x \to {0^ - }} x\,\sqrt {\frac{{2x}}{{x + 3}}} = \not \exists \\
\mathop {\lim }\limits_{x \to {0^ + }} x\,\sqrt {\frac{{2x}}{{x + 3}}} = 0\,\sqrt {\frac{0}{3}} = 0
\end{array} \right.\]

- Studio del segno
\[\left[ \begin{array}{l}
f(x) \ge 0\;\; \to \;\;x\,\sqrt {\frac{{2x}}{{x + 3}}} \ge 0\;\; \to \;\;x \ge 0\\
f(x) < 0\;\; \to \;\;x\,\sqrt {\frac{{2x}}{{x + 3}}} < 0\;\; \to \;\;x < - 3
\end{array} \right.\]

- Asintoti Orizzontali e Obliqui
\[x \to \infty \;\;\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } x\,\sqrt {\frac{{2x}}{{x + 3}}} = \mathop {\lim }\limits_{x \to + \infty } x\,\sqrt {\frac{{2x}}{{x(1 + \frac{3}{x})}}} = \mathop {\lim }\limits_{x \to + \infty } x\,\sqrt {\frac{2}{{1 + \frac{3}{x}}}} = + \infty \,\sqrt {\frac{2}{{1 + 0}}} = + \infty \\
\mathop {\lim }\limits_{x \to - \infty } x\,\sqrt {\frac{{2x}}{{x + 3}}} = \mathop {\lim }\limits_{x \to - \infty } x\,\sqrt {\frac{{2x}}{{x(1 + \frac{3}{x})}}} = \mathop {\lim }\limits_{x \to - \infty } x\,\sqrt {\frac{2}{{1 + \frac{3}{x}}}} = - \infty \,\sqrt {\frac{2}{{1 + 0}}} = - \infty
\end{array} \right.\]
\[y = mx + q\]
\[m = \mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{x\,\sqrt {\frac{{2x}}{{x + 3}}} }}{x} = \mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{2x}}{{x + 3}}} = \mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{2x}}{{x(1 + \frac{3}{x})}}} = \]
\[ = \mathop {\lim }\limits_{x \to \infty } x\,\sqrt {\frac{2}{{1 + \frac{3}{x}}}} = \sqrt 2 \]
\[q = \mathop {\lim }\limits_{x \to \infty } \,[f(x) - mx] = \mathop {\lim }\limits_{x \to \infty } \,[x\,\sqrt {\frac{{2x}}{{x + 3}}} - \sqrt 2 x] = \mathop {\lim }\limits_{x \to \infty } \,x\,[\sqrt {\frac{{2x}}{{x + 3}}} - \sqrt 2 ] = \]
\[ = \mathop {\lim }\limits_{x \to \infty } \,x\,[\sqrt {\frac{{2x}}{{x + 3}}} - \sqrt 2 ] = \mathop {\lim }\limits_{x \to \infty } \,x\,[\frac{{(\sqrt {\frac{{2x}}{{x + 3}}} - \sqrt 2 )(\sqrt {\frac{{2x}}{{x + 3}}} + \sqrt 2 )}}{{\sqrt {\frac{{2x}}{{x + 3}}} + \sqrt 2 }}] = \]
\[ = \mathop {\lim }\limits_{x \to \infty } \,x\,[\frac{{\frac{{2x}}{{x + 3}} - 2}}{{\sqrt {\frac{{2x}}{{x + 3}}} + \sqrt 2 }}] = \mathop {\lim }\limits_{x \to \infty } \,x\,[\frac{{\frac{{2x - 2x - 6}}{{x + 3}}}}{{\sqrt {\frac{{2x}}{{x + 3}}} + \sqrt 2 }}] = \mathop {\lim }\limits_{x \to \infty } \,x\,[\frac{{ - 6x}}{{(x + 3)(\sqrt {\frac{{2x}}{{x + 3}}} + \sqrt 2 )}}] = \]
\[ = \mathop {\lim }\limits_{x \to \infty } \,\frac{{ - 6x}}{{x + 3}} \cdot \mathop {\lim }\limits_{x \to \infty } \,\frac{1}{{(\sqrt {\frac{{2x}}{{x + 3}}} + \sqrt 2 )}} = - 6 \cdot \frac{1}{{2\sqrt 2 }} = \frac{{ - 3}}{{\sqrt 2 }} = - \frac{3}{2}\sqrt 2 \]
\[y = \sqrt 2 \,x - \frac{3}{2}\sqrt 2 \;\;\;\;({\rm{Asintoto}}\;{\rm{Obliquo)}}\]

- Derivata prima (monotonia/max/min)
\[f'(x) = 1 \cdot \sqrt {\frac{{2x}}{{x + 3}}} + x\frac{1}{{2\sqrt {\frac{{2x}}{{x + 3}}} }}[\frac{{2(x + 3) - 2x}}{{{{(x + 3)}^2}}}] = \]
\[ = \sqrt {\frac{{2x}}{{x + 3}}} + \frac{{6x}}{{2{{(x + 3)}^2}\sqrt {\frac{{2x}}{{x + 3}}} }} = \sqrt {\frac{{2x}}{{x + 3}}} + \frac{{3x}}{{{{(x + 3)}^2}\frac{{2x}}{{x + 3}}}}\sqrt {\frac{{2x}}{{x + 3}}} = \]
\[ = \sqrt {\frac{{2x}}{{x + 3}}} \;[1 + \frac{{3x}}{{{{(x + 3)}^2}\frac{{2x}}{{x + 3}}}}] = \sqrt {\frac{{2x}}{{x + 3}}} \;[1 + \frac{{3x}}{{2x(x + 3)}}] = \]
\[ = \sqrt {\frac{{2x}}{{x + 3}}} \;[1 + \frac{3}{{2(x + 3)}}] = \sqrt {\frac{{2x}}{{x + 3}}} \;[\frac{{2(x + 3) + 3}}{{2(x + 3)}}] = \]
\[ = \sqrt {\frac{{2x}}{{x + 3}}} \;[\frac{{2x + 9}}{{2(x + 3)}}]\]
\[{\rm{Max}}\;{\rm{Rel: }}x = - \frac{9}{2}\]

- Derivata seconda (curvature/concavità/convessità/flessi)
Lascio a te i calcoli della derivata seconda e l'eventuale fattibilità del suo studio studio del segno

- Grafico in allegato


Saluti.