durchschnitt des ersten und dritten Quartils {0.42, 0.52, 0.58, 0.62}

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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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range x³−2x²−3xx²−4

Solution

−∞ <f(x)<∞

Interval Notation

(−∞ , ∞ )

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Solution steps

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Function range definition

The set of values of the dependent variable for which a function is defined

Find the minimum and maximum value in each defined interval and unite the results

Domain of x³−2x²−3xx²−4 : x<−2 or −2<x<2 or x>2

Extreme Points of x³−2x²−3xx²−4 : Maximum(−3.54391…, −6.89260…), Minimum(−0.57240…, −0.23809…)

Find the range for the interval −∞ <x<−2: −∞ <f(x)≤ −6.89260…

Find the range for the interval −2<x<2: −0.23809…≤ f(x)<∞

Find the range for the interval 2<x<∞ : −∞ <f(x)<∞

Combine the ranges of all domain intervals to obtain the function range

f(x)≤ −6.89260… or f(x)≥ −0.23809… or −∞ <f(x)<∞

−∞ <f(x)<∞

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domain x³−2x²−3xx²−4

intercepts x³−2x²−3xx²−4

parity x³−2x²−3xx²−4

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

durchschnitt des ersten und dritten Quartils {0.42, 0.52, 0.58, 0.62}

Solution

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