amplitude of 1-cos^2(x)

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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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domínio 2sin⁵(x)

Solution

−∞ <x<∞

Interval Notation

(−∞ , ∞ )

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Function Inverse definition

A function g is the inverse of function f if for y=f(x), x=g(y)

y=(x4 )²

Replace x with y

x=(y4 )²

Solve for y, x=(y4 )²

4√x, −4√x

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intervalo 2sin⁵(x)

periodicidade 2sin⁵(x)

amplitude 2sin⁵(x)

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

amplitude of 1-cos^2(x)

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amplitude of 1-cos^2(x)

Solution

Solution steps

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