prove that 4ln((sqrt(2))/2)=-2ln(2)

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(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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Steps Examples

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gamme y=x²+x+1x

Solution

f(x)≤ −1 or f(x)≥ 3

Interval Notation

(−∞ , −1]∪ [3, ∞ )

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Solution steps

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Définition de la plage de fonction

L'ensemble des valeurs de la variable dépendante pour lesquelles une fonction est définie

Récrire comme

x²+x+1x =y

Multiplier les deux côtés par x

x²+x+1=yx

La plage est l'ensemble des y pour lesquels le discriminant est supérieur ou égal à zéro

Discriminant noté x²+x+1=yx: y²−2y−3

y²−2y−3≥ 0 : y≤ −1 or y≥ 3

Vérifier si les points d'extrémités de la plage d'intervalle sont inclus

y=−1 : Inclus

y=3 : Inclus

Par conséquent la plage est

f(x)≤ −1 or f(x)≥ 3

Solve instead

domaine indéfini x²+x+1x

inverser x²+x+1x

points d'intersection x²+x+1x

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

prove that 4ln((sqrt(2))/2)=-2ln(2)

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