Question

19.2 определите область определённой функции:

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Answer 1

Ответ:

$1)\ \ y=\sqrt{(5-x)(2x-7)}\\\\OOF:\ \ \ (5-x)(2x-7)\geq 0\ \ \ \to \ \ \ (x-5)(2x-7)\leq 0\ \ ,\\\\x-5=0\ \ \to \ \ x=5\\\\2x-7=0\ \ \to \ \ \ x=3,5\\\\znaki:\ \ +++(3,5)---(5)+++\\\\x\in [\ 3,5\ ;\ 5\ ]\\\\\\2)\ \ y=\sqrt{(5+3x)(2x-2,4)}\\\\OOF:\ \ (5+3x)(2x-2,4)\geq 0\\\\5+3x=0\ \ \to \ \ \ x=-\dfrac{5}{3}=-1\dfrac{2}{3}\\\\2x-2,4=0\ \ \to \ \ \ x=1,2=1\dfrac{1}{5}\\\\znaki:\ \ +++(-\frac{5}{3})---(1,2)+++\\\\x\in (-\infty ;-1\frac{2}{3}\ ]\cup [\ 1\frac{1}{5}\ ;+\infty \, )$

$3)\ \ y=\sqrt{(4+x)(5-2x)}\\\\OOF:\ \ (4+x)(5-2x)\geq 0\ \ ,\\\\4+x=0\ \ \to \ \ \ x=-4\\\\5-2x=0\ \ \to \ \ x=2,5\\\\znaki:\ \ \ ---(2,5)+++(4)---\\\\x\in [\ 2,5\ ;\ 4\ ]\\\\\\4)\ \ y=\sqrt{-2(1-x)(2x+5)}\\\\OOF:\ \ -2(1-x)(2x+5)\geq 0\ \ \to \ \ \ 2(x-1)(2x+5)\geq 0\\\\x-1=0\ \ \to \ \ x=1\\\\2x+5=0\ \ \to \ \ \ x=-2,5\\\\znaki:\ \ +++(-2,5)---(\, 1\, )+++\\\\x\in (-\infty ;-2,5\, ]\cup [\ 1\ ;+\infty \, )$