Question

Найти область определения функции

Answer 1

Ответ:

$y=\sqrt{(x^2-1)\, log_{1/3}(3-x)}\\\\OOF:\ \ (x^2-1)\, log_{1/3}(3-x)\geq 0\ \ \Rightarrow \\\\\\\left\{\begin{array}{l}x^2-1\geq 0\\log_{1/3}(3-x)\geq 0\end{array}\right\ \ \ \ ili\ \ \ \ \ \left\{\begin{array}{l}x^2-1\leq 0\\log_{1/3}(3-x)\leq 0\end{array}\right\\\\\\\left\{\begin{array}{l}(x-1)(x+1)\geq 0\\0<3-x\leq 1\end{array}\right\ \ \ \ ili\ \ \ \ \left\{\begin{array}{l}(x-1)(x+1)\leq 0\\3-x\geq 1\end{array}\right$

$\left\{\begin{array}{l}x\in (-\infty \, ;\, -1\ ]\cup [\ 1\, ;+\infty \, )\\2\leq x<3\end{array}\right\ \ \ \ ili\ \ \ \ \left\{\begin{array}{l}x\in [-1\, ;\, 1\, ]\\x\leq 2\end{array}\right\\\\\\x\in [\ 2\, ;\, 3\, )\qquad \qquad \qquad \qquad \quad \ \ \ \ ili\ \ \ \ \ \ \ x\in [-1\, ;\, 1\ ]\\\\Otvet:\ \ x\in [-1\, ;\, 1\ ]\cup [\ 2\, ;\ 3\, )\ .$