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領域 3a^2+1-(4+2a^2)

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

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domain 3a²+1−(4+2a²)

Solution

−∞ <a<∞

Interval Notation

(−∞ , ∞ )

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Domain definition

The domain of a function is the set of input or argument values for which the function is real and defined

The function has no undefined points nor domain constraints. Therefore, the domain is

−∞ <a<∞

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

領域 3a^2+1-(4+2a^2)

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領域 3a^2+1-(4+2a^2)

Solution

Solution steps

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