라플라스변환 cos(3t)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

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∂ ∂ x (√x⁹−y¹⁵)

Solution

9x⁸2√x⁹−y¹⁵

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Solution steps

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함수 범위 정의

함수가 정의되는 종속 변수의 값 집합

정의된 각 간격에서 최소값과 최대값을 찾고 결과를 통합합니다

도메인 5x−x²x⁴−25x² : x<−5 or −5<x<0 or 0<x<5 or x>5

5x−x²x⁴−25x² 단순화하세요: −1x(x+5)

극한점 −1x(x+5) : 극히 적은 양(−52 , 425 )

간격의 범위를 찾습니다 −∞ <x<−5: −∞ <f(x)<0

간격의 범위를 찾습니다 −5<x<0: 425 ≤ f(x)<∞

간격의 범위를 찾습니다 0<x<5: −∞ <f(x)<−150

간격의 범위를 찾습니다 5<x<∞ : −150 <f(x)<0

모든 도메인 간격의 범위를 결합하여 함수 범위를 얻습니다

f(x)<0 or f(x)≥ 425 or f(x)<−150 or −150 <f(x)<0

f(x)<0 or f(x)≥ 425

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안녕, 질문하다 ∂ ∂ x (√x⁹−y¹⁵) 또는 아래 옵션 중에서 선택하세요

함수의 범위를 찾는 방법을 설명하세요

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

라플라스변환 cos(3t)

Solution

Solution steps

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