Ciao, ecco i passi essenziali ...
\[f(x) = x\,{e^{\frac{1}{{1 + x}}}}\]
- Dominio e studio dei punti singolari
\[{D_f} = R - \{ - 1\} \]
\[{x_0} = - 1\;\left\{ \begin{array}{l}
f( - 1) = \not \exists \\
\mathop {\lim }\limits_{x \to - {1^ - }} x\,{e^{\frac{1}{{1 + x}}}} = - 1\,{e^{\frac{1}{{{0^ - }}}}} = - 1\,{e^{ - \infty }} = 0\\
\mathop {\lim }\limits_{x \to - {1^ + }} x\,{e^{\frac{1}{{1 + x}}}} = - 1\,{e^{\frac{1}{{{0^ + }}}}} = - 1\,{e^{ + \infty }} = - \infty \;(A.V.)
\end{array} \right.\]
- Studio del segno e intersezione con assi
\[\left[ \begin{array}{l}
f(x) > 0\;\;\; \to \;\;\;x > 0\\
f(x) = 0\;\;\; \to \;\;\;x = 0\\
f(x) < 0\;\;\; \to \;\;\;x < 0
\end{array} \right.\]
\[\left[ \begin{array}{l}
\cap \;asse\;x\;\; \to \,\;x = 0\\
\cap \;asse\;y\;\; \to \;f(0) = 0
\end{array} \right.\]
- Asintoti Orizzontali e Obliqui
\[\left[ \begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } x\,{e^{\frac{1}{{1 + x}}}} = + \infty \,{e^0} = + \infty \\
\mathop {\lim }\limits_{x \to - \infty } x\,{e^{\frac{1}{{1 + x}}}} = - \infty \,{e^0} = - \infty
\end{array} \right.\;\;NO\;A.Oz.\]
\[y = mx + q\]
\[m = \mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{x\,{e^{\frac{1}{{1 + x}}}}}}{x} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{{1 + x}}}} = {e^0} = 1\]
\[q = \mathop {\lim }\limits_{x \to \infty } \,[f(x) - mx] = \mathop {\lim }\limits_{x \to \infty } \,[x\,{e^{\frac{1}{{1 + x}}}} - x] = \mathop {\lim }\limits_{x \to \infty } x\,[{e^{\frac{1}{{1 + x}}}} - 1] = \]
\[ = \mathop {\lim }\limits_{x \to \infty } x\,[\frac{{{e^{\frac{1}{{1 + x}}}} - 1}}{{\frac{1}{{1 + x}}}}]\frac{1}{{1 + x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^{\frac{1}{{1 + x}}}} - 1}}{{\frac{1}{{1 + x}}}} \cdot \mathop {\lim }\limits_{x \to \infty } \frac{x}{{1 + x}} = 1 \cdot 1 = 1\]
\[y = x + 1\;\;(A.Ob.)\]
- Studio derivata prima (crescenza/decrescenza/max/min)
\[f'(x) = 1\,{e^{\frac{1}{{1 + x}}}} + x\,{e^{\frac{1}{{1 + x}}}}( - \frac{1}{{{{(1 + x)}^2}}}) = {e^{\frac{1}{{1 + x}}}}\,[1 - \frac{x}{{{{(1 + x)}^2}}}] = \]
\[ = {e^{\frac{1}{{1 + x}}}}\,[\frac{{{{(1 + x)}^2} - x}}{{{{(1 + x)}^2}}}] = {e^{\frac{1}{{1 + x}}}}\,[\frac{{{x^2} + x + 1}}{{{{(1 + x)}^2}}}] > 0\;\;\; \to \;\;\;\forall x \in {D_f}\;\;(strett.\,cresc.)\]
- Studio curvature (concavità/convessità/flessi) : lascio a te il calcolo e l'eventuale studio del segno della derivata seconda!
- Grafico della funzione : in allegato
Saluti
