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

Solutions > One-Step Division Equation Calculator >

pendiente 3x+26

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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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Steps Examples

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puntos extremos f(x)=sin(3x)

Solution

Máximo(π6 +2π3 n, 1), Mínimo(π2 +2π3 n, −1)

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Solve by:

Encontrar utilizando el criterio de la primera derivada

Encontrar utilizando el criterio de la primera derivada

Encontrar utilizando el criterio de la segunda derivada

One step at a time

Definicion del criterio de la primera derivada

Suponer que x=c es un punto crítico de f(x) entonces,
Si f ′(x)>0 a la izquierda de x=c y f ′(x)<0 a la derecha de x=c entonces x=c es un máximo local
Si f ′(x)<0 a la izquierda de x=c y f ′(x)> 0 a la derecha de x=c entonces x=c es un mínimo local.
Si f ′(x) tiene el mismo signo en ambos lados de x=c entonces x=c no es ni un máximo local ni un mínimo local.

Encontrar los puntos criticos: x=π6 +2π3 n, x=π2 +2π3 n

Dominio de sin(3x) : −∞ <x<∞

Encontrar intervalos: Creciente:2π3 n≤ x<π6 +2π3 n, Decreciente:π6 +2π3 n<x<π2 +2π3 n, Creciente:π2 +2π3 n<x<2π3 +2π3 n

Sustituir el punto extremo x=π6 +2π3 n en sin(3x) ⇒ y=1

Máximo(π6 +2π3 n, 1)

Sustituir el punto extremo x=π2 +2π3 n en sin(3x) ⇒ y=−1

Mínimo(π2 +2π3 n, −1)

Máximo(π6 +2π3 n, 1), Mínimo(π2 +2π3 n, −1)

Solve instead

domínio sin(3x)

rango sin(3x)

periodicidad sin(3x)

One-Step Division Equation Examples

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