четность f(x)=cos(2x+5)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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∫

ddx

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

simplify solve for inverse tangent line

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domain x³−2x²−3xx²−4

Solution

x<−2 or −2<x<2 or x>2

Interval Notation

(−∞ , −2)∪(−2, 2)∪(2, ∞ )

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Area under a polar curve definition

The area under a polar curve given by r = f(θ) between the radial lines θ = a and θ = b is
A=12 ∫_a^br²dθ

Identify the interval: (0, π2 ), (3π2 , 2π)

Apply the area formula: 12 · ∫₀^π24cos(θ)dθ+12 · ∫_3π2^2π4cos(θ)dθ

Solve 12 ∫₀^π24cos(θ)dθ+12 ∫_3π2^2π4cos(θ)dθ: 4

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range 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`

четность f(x)=cos(2x+5)

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