points of d'intersection -1/2 x+1/2

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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

simplify solve for expand factor rationalize

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gamme 20(x−3)(x+4)(x−3)²(x−5)

Solution

f(x)≤ −30√7−80 or 30√7−80≤ f(x)<0 or f(x)>0

Interval Notation

(−∞ , −30√7−80]∪ [30√7−80, 0)∪(0, ∞ )

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Définition de la plage de fonction

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

La plage de la fonction est le domaine de définition combiné des fonctions inverses

Trouver la(les) fonction(s) inverse(e) de : 20(x−3)(x+4)(x−3)²(x−5)

Fonction inverse de 20(x−3)(x+4)(x−3)²(x−5) : 4x+10+√x²+160x+100x , 4x+10−√x²+160x+100x

4x+10+√x²+160x+100x , 4x+10−√x²+160x+100x

Trouver le domaine de chaque fonction inverse

Domaine de 4x+10+√x²+160x+100x : x≤ −30√7−80 or 30√7−80≤ x<0 or x>0

Domaine de 4x+10−√x²+160x+100x : x≤ −30√7−80 or 30√7−80≤ x<0 or x>0

Réunir les intervalles

f(x)≤ −30√7−80 or 30√7−80≤ f(x)<0 or f(x)>0

Solve instead

domaine indéfini 20(x−3)(x+4)(x−3)²(x−5)

inverser 20(x−3)(x+4)(x−3)²(x−5)

points d'intersection 20(x−3)(x+4)(x−3)²(x−5)

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

points of d'intersection -1/2 x+1/2

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