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f(x)=x^3x^8

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

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ddx

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

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gamme x²+3x−4x²+x−12

Solution

f(x)<57 or 57 <f(x)<1 or f(x)>1

Interval Notation

(−∞ , 57 )∪(57 , 1)∪(1, ∞ )

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Définition de la plage de fonction

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

Trouver la valeur minimale et la valeur maximale dans chaque intervalle défini et unir les résultats

Domaine de x²+3x−4x²+x−12 : x<−4 or −4<x<3 or x>3

Simplifier x²+3x−4x²+x−12 : x−1x−3

Points extrêmes de x−1x−3 : Aucun

Trouver la plage pour l'intervalle −∞ <x<−4: 57 <f(x)<1

Trouver la plage pour l'intervalle −4<x<3: −∞ <f(x)<57

Trouver la plage pour l'intervalle 3<x<∞ : 1<f(x)<∞

Combiner les plages de tous les intervalles du domaine pour obtenir la plage de la fonction

57 <f(x)<1 or f(x)<57 or f(x)>1

f(x)<57 or 57 <f(x)<1 or f(x)>1

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domaine indéfini x²+3x−4x²+x−12

inverser x²+3x−4x²+x−12

points d'intersection x²+3x−4x²+x−12

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

f(x)=x^3x^8

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