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

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

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ddx

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simplify solve for inverse tangent line

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f(x)=(x+1)(x−1)

Solution

Domain: −∞ <x<∞

Range: f(x)≥ −1

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

Vertex: Minimum (0, −1)

Inverse: √x+1, −√x+1

Interval Notation

Domain: (−∞ , ∞ )

Range: [−1, ∞ )

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Domain of (x+1)(x−1) : −∞ <x<∞

Range of (x+1)(x−1): f(x)≥ −1

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

Vertex of (x+1)(x−1): Minimum (0, −1)

Inverse of (x+1)(x−1): √x+1, −√x+1

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ddx ((x+1)(x−1))

parity (x+1)(x−1)

symmetry (x+1)(x−1)

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

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

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