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vértices 3a^2+1-(4+2a^2)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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∞

∫

ddx

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

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range 2sin⁵(x)

Solution

−2≤ f(x)≤ 2

Interval Notation

[−2, 2]

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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 2sin⁵(x) : −∞ <x<∞

Extreme Points of 2sin⁵(x): Minimum(2πn, 0), Maximum(π2 +2πn, 2), Saddle(π+2πn, 0), Minimum(3π2 +2πn, −2)

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

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

−2≤ f(x)≤ 2

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domain 2sin⁵(x)

periodicity 2sin⁵(x)

amplitude 2sin⁵(x)

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

vértices 3a^2+1-(4+2a^2)

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vértices 3a^2+1-(4+2a^2)

Solution

Solution steps

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