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диапазон 1/(x^2)+1/x

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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диапазон 1x²

Solution

f(x)>0

Interval Notation

(0, ∞ )

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Определение диапазона функций

Набор значений зависимой переменной, для которой определена функция

Перепишите как

1x² =y

Умножьте обе части на x²

1=yx²

Диапазон - это набор значений y, для которых дискриминант больше или равен нулю

Дискриминант 1=yx²: 4y

4y≥ 0 : y≥ 0

Проверьте, включены ли конечные точки интервала диапазона

y=0 : Исключен

Поэтому диапазон равен

f(x)>0

Solve instead

домен 1x²

обратная 1x²

перехваты 1x²

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

диапазон 1/(x^2)+1/x

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