f(x)=(x^2+1)(x^2-1)

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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x^☐

√☐

^□√☐

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
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simplify solve for inverse tangent line

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f(x)=(x²+1)(x²−1)

Solution

Domain: −∞ <x<∞

Range: f(x)≥ −1

X Intercepts: (−1, 0), (1, 0), Y Intercepts: (0, −1)

Asymptotes: None

Extreme Points: Minimum(0, −1)

Interval Notation

Domain: (−∞ , ∞ )

Range: [−1, ∞ )

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Domain of (x²+1)(x²−1) : −∞ <x<∞

Range of (x²+1)(x²−1): f(x)≥ −1

Axis interception points of (x²+1)(x²−1): X Intercepts: (−1, 0), (1, 0), Y Intercepts: (0, −1)

Asymptotes of (x²+1)(x²−1): None

Extreme Points of (x²+1)(x²−1): Minimum(0, −1)

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ddx ((x²+1)(x²−1))

inverse (x²+1)(x²−1)

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

f(x)=(x^2+1)(x^2-1)

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