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

Solutions > Functions Domain of Inverse Calculator >

inverse of sqrt(x-4)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

π

θ

∞

∫

ddx

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

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domain √x+1

Solution

x≥ −1

Interval Notation

[−1, ∞ )

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Solve by:

Encontrar utilizando o critério da segunda derivada

Encontrar utilizando o critério da segunda derivada

Encontrar utilizando o critério da primeira derivada

One step at a time

Definição do critério da segunda derivada

Suponha que x=c é um ponto crítico de f ′(c) tal que f ′(c)=0
e que f ′′(x) é contínuo em uma regiãoao redor de x=c. Então,
Sef ′′(c)<0 então x=c é um máximo local.
Se f ′′(c)>0 então x=c é um mínimo local.
Se f ′′(c)=0 então o teste falhou e x=c pode ser um máximo local, mínimo local ou nenhum dos dois .

f ′(x)=2x−6

Find intervals: Decrescente:−∞ <x<3, Crescente:3<x<∞

f ′′(x)=2

Verificar o sinal de f ′′(x)=2 em cada ponto crítico

Verificar o ponto crítico x=3: Mínimo(3, −1)

Mínimo(3, −1)

Solve instead

inverse √x+1

intercepts √x+1

Functions Domain of Inverse Examples

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