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доказывать (1-sin(x))/(1+sin(x))=(sec(x)-tan(x))^2

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диапазон xx²−6x+8

Solution

f(x)≤ −√2−32 or f(x)≥√2−32

Interval Notation

(−∞ , −√2−32 ]∪ [√2−32 , ∞ )

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Solution steps

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Определение диапазона функций

Набор значений зависимой переменной, для которой определена функция

Перепишите как

xx²−6x+8 =y

Умножьте обе части на x²−6x+8

x=y(x²−6x+8)

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

Дискриминант x=y(x²−6x+8): 4y²+12y+1

4y²+12y+1≥ 0 : y≤ −√2−32 or y≥ √2−32

Проверьте, включены ли конечные точки интервала диапазона

y=−√2−32 : Включен

y=√2−32 : Включен

Поэтому диапазон равен

f(x)≤ −√2−32 or f(x)≥ √2−32

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домен xx²−6x+8

обратная xx²−6x+8

перехваты xx²−6x+8

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

доказывать (1-sin(x))/(1+sin(x))=(sec(x)-tan(x))^2

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