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area x^2+(y+3)^2=16

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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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inversa √x²+1

Solution

√x²−1, −√x²−1

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Definição de função inversa

A função g é o inverso da função f se para y=f(x), x=g(y)

y=√x²+1

Substitua x por y

x=√y²+1

Resolver para y, x=√y²+1

√x²−1, −√x²−1

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domínio √x²+1

intervalo √x²+1

interceptos √x²+1

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

area x^2+(y+3)^2=16

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area x^2+(y+3)^2=16

Solution

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