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vértices y=2x^2

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

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ddx

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

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inverse (x²+1)(x²−1)

Solution

⁴√x+1, −⁴√x+1

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Function Inverse definition

A function g is the inverse of function f if for y=f(x), x=g(y)

y=(x²+1)(x²−1)

Replace x with y

x=(y²+1)(y²−1)

Solve for y, x=(y²+1)(y²−1)

⁴√x+1, −⁴√x+1

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domain (x²+1)(x²−1)

range (x²+1)(x²−1)

intercepts (x²+1)(x²−1)

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

vértices y=2x^2

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vértices y=2x^2

Solution

Solution steps

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