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Sa se determine domeniul de definitie al expresiei E= log x+1(3x^2-12)

Răspuns :

[tex]\left\{ \begin{array}{ll} x+1 \ \textgreater \ 0\\ x+1 \neq 1 \\ 3x^2-12\ \textgreater \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} x\ \textgreater \ -1\\ x\neq0\\ 3x^2\ \textgreater \ 12 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} x\ \textgreater \ -1\\ x\neq0\\ x^2\ \textgreater \ 4\Big|\sqrt\boxed~} \end{array} \right \Rightarrow \\ \\ [/tex]

[tex]\Rightarrow \left\{ \begin{array}{ll} x\ \textgreater \ -1\\ x\neq0\\ \sqrt{x^2}\ \textgreater \ \sqrt4 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} x\ \textgreater \ -1\\ x\neq0\\ |x| \ \textgreater \ 2 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} x\in (-1,\infty)\\ x\in (-\infty,0)\cup (0,+\infty)\\ x\in (-\infty,-2)\cup(2,+\infty) \end{array} \right \Rightarrow \\ \\\\[/tex]

[tex] -\infty~~~~~~~~~-2~~~~~~~~~-1~~~~~~~~~~0~~~~~~~~~~2~~~~~~~~~~~+\infty\\ -$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$-$- \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(-------------$-~) \\ ~~(--------------)(----------)\\ ~~(------)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(------$~)$ \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(||||||||||||||||||||||)\\ \Rightarrow \boxed{D=(2,+\infty)}[/tex]