Studio di funzione trigonometrica

[plaltz168]

vorrei sapere come si fa a svolgere lo studio della seguente funzione arctan[(|2x+1|-x)/(|3x-2|-1)]

[Matlurker]

[math]\arctan{\frac{|2x+1|-x}{|3x-2|-1}}
\\\text{Dominio:}
\\\begin{cases} -3x+2-1 \neq 0 \Rightarrow x \neq \frac{1}{3}
\\\ 3x-2-1 \neq 0 \Rightarrow x \neq 1 \end{cases}
\\\boxed{\mathcal{D}: \mathbb{R} \setminus \left \{\frac{1}{3};1 \right \}}
\\\text{Forma alternativa:}
\\f(x)=\begin{cases}
\arctan{\frac{3x+1}{3x-1}}\quad \text{ per } x \in \left(-\infty,-\frac{1}{2}\right)
\\\arctan{\frac{x+1}{1-3x}}\quad \text{ per } x \in \left[-\frac{1}{2},\frac{1}{3}\right) \cup \left(\frac{1}{3}, \frac{2}{3}\right)
\\\arctan{\frac{x+1}{3x-3}}\quad \text{ per } x \in \left[\frac{2}{3}, 1\right) \cup \left(1, +\infty\right)
\\\end{cases}
\\\text{Asintoti:}
\\\begin{cases} \lim_{x \to \frac{1^-}{3}}{\arctan{\frac{x+1}{1-3x}}}=+\infty
\\\lim_{x \to \frac{1^+}{3}}{\arctan{\frac{x+1}{1-3x}}}=-\infty
\end{cases} \Rightarrow \boxed{x=\frac{1}{3}}
\\\begin{cases}\lim_{x \to {1^-}}{\arctan{\frac{x+1}{3x-3}}}=-\infty
\\\lim_{x \to {1^+}}{\arctan{\frac{x+1}{3x-3}}}=+\infty
\end{cases}\Rightarrow \boxed{x=1}
\\\lim_{x \to {-\infty}}{\arctan{\frac{3x+1}{3x-1}}}=\arctan{1}=\frac{\pi}{4} \Rightarrow \boxed{y=\frac{\pi}{4}}
\\\lim_{x \to {+\infty}}{\arctan{\frac{x+1}{3x-3}}}=\arctan{\frac{1}{3}} \Rightarrow \boxed{y=\arctan{\frac{1}{3}}}
\\\lim_{x \to \infty}{\frac{f(x)}{x}}=m=0 \Rightarrow \boxed{\text{Nessun asintoto obliquo}}
\\\text{Derivata}
\\f'(x)=\begin{cases}
\frac{1}{1+\left(\frac{3x+1}{3x-1}\right)^2} \cdot \frac{9x-3-9x-9}{\left(3x-1\right)^2} \quad \text{ per } x \in \left(-\infty,-\frac{1}{2}\right)
\\\frac{1}{1+\left(\frac{x+1}{1-3x}\right)^2} \cdot \frac{1-3x+3x+3}{\left(1-3x\right)^2} \quad \text{ per } x \in \left[-\frac{1}{2},\frac{1}{3}\right) \cup \left(\frac{1}{3}, \frac{2}{3}\right)
\\\frac{1}{1+\left(\frac{x+1}{3x-3}\right)^2} \cdot \frac{3x-3-3x-3}{\left(3x-3\right)^2} \quad \text{ per } x \in \left[\frac{2}{3}, 1\right) \cup \left(1, +\infty\right)
\end{cases} \Rightarrow
\\\Rightarrow f'(x)=\begin{cases}
\frac{\left(3x-1\right)^2}{18x^2+2} \cdot \frac{-12}{\left(3x-1\right)^2} \quad \text{ per } x \in \left(-\infty,-\frac{1}{2}\right)
\\\frac{\left(1-3x\right)^2}{10x^2-4x+2} \cdot \frac{4}{\left(1-3x\right)^2} \quad \text{ per } x \in \left[-\frac{1}{2},\frac{1}{3}\right) \cup \left(\frac{1}{3}, \frac{2}{3}\right)
\\\frac{\left(3x-3\right)^2}{9x^2-18x+9+x^2+2x+1} \cdot \frac{-6}{\left(3x-3\right)^2} \quad \text{ per } x \in \left[\frac{2}{3}, 1\right) \cup \left(1, +\infty\right)
\end{cases}
\\\Rightarrow f'(x)=\begin{cases}
\frac{-3}{9x^2+1} \quad \text{ per } x \in \left(-\infty,-\frac{1}{2}\right)
\\\frac{2}{5x^2-2x+1} \quad \text{ per } x \in \left[-\frac{1}{2},\frac{1}{3}\right) \cup \left(\frac{1}{3}, \frac{2}{3}\right)
\\\frac{-3}{5x^2-8x+5} \quad \text{ per } x \in \left[\frac{2}{3}, 1\right) \cup \left(1, +\infty\right)
\end{cases}
\\\Delta_1