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absolute extreme points of f(x)=sqrt(x+3)

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

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

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área y=x²−1, (0, 1)

Solution

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0.66666…

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Solve by:

Encontrar el vértice con el promedio de los ceros

Encontrar el vértice utilizando la forma polinómica

Encontrar el vértice utilizando la forma parábola

Encontrar el vértice utilizando la forma canónica

One step at a time

y=(2x−1)(5x−6)

Ecuación de parábola en forma factorizada

El vértice de una parábola abierta arriba abajo de la forma y=a(x−m)(x−n) es el promedio de sus ceros x_v=m+n2

y=(2x−1)(5x−6)

Los parámetros de la parábola son:

a=10, m=12 , n=65

x_v=m+n2

x_v=(12 )+(65 )2

Simplificar (12 )+(65 )2 : 1720

x_v=1720

Ingresar x_v=1720 para encontrar el valor y_v

y_v=−4940

Por lo tanto, el vertice de la parabola es

(1720 , −4940 )

Si a<0, entonces el vertice es un valor máximo
Si a>0, entonces el vertice es un valor minimo
a=10

Mínimo (1720 , −4940 )

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

absolute extreme points of f(x)=sqrt(x+3)

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