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vertices 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^☐

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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 (2x²+6)(2x²−6)

Solution

f(x)≥ −36

Interval Notation

[−36, ∞ )

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Function range definition

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

The function range is the combined domain of the inverse functions

Find the inverse function(s) of: (2x²+6)(2x²−6)

Inverse of (2x²+6)(2x²−6): √√x+362 , −√√x+362 , √−√x+362 , −√−√x+362

√√x+362 , −√√x+362 , √−√x+362 , −√−√x+362

Find the domain of each inverse function

Domain of √√x+362 : x≥ −36

Domain of −√√x+362 : x≥ −36

Domain of √−√x+362 : x=−36

Domain of −√−√x+362 : x=−36

Combine the intervals

f(x)≥ −36

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domain (2x²+6)(2x²−6)

inverse (2x²+6)(2x²−6)

intercepts (2x²+6)(2x²−6)

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

vertices 3a^2+1-(4+2a^2)

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vertices 3a^2+1-(4+2a^2)

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