interceptos (x^2+1)(x^2-1)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

simplify solve for partial fractions long division line

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inversa (z5 )²

Solution

5√z, −5√z

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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=(z5 )²

Substitua z por y

z=(y5 )²

Resolver para y, z=(y5 )²

5√z, −5√z

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domínio (z5 )²

intervalo (z5 )²

vértices (z5 )²

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

`x`²	`x`<−2
ln(`x`+5)	`x`≥−2

sin(`x`)	`x`<0
2`x`	`x`≥0

√`x`+2	`x`<1
`xx`²−6`x`+8	`x`≥1

√`x`²−4	`x`<0
`x`³−`x`	`x`≥0

interceptos (x^2+1)(x^2-1)

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interceptos (x^2+1)(x^2-1)

Solution

Solution steps

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