expandir (x^2+5)(x^2-5)

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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

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inverse (z−8)²z²−64

Solution

−8−8zz−1

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Definição de imagem de função

O conjunto de valores da variável dependente para a qual uma função é definida

Encontrar o valor mínimo e máximo em cada intervalo definido e unificar os resultados

Domínio de 5x−x²x⁴−25x² : x<−5 or −5<x<0 or 0<x<5 or x>5

Simplificar 5x−x²x⁴−25x² : −1x(x+5)

Pontos extremos de −1x(x+5) : Mínimo(−52 , 425 )

Encontrar a imagem para o intervalo −∞ <x<−5: −∞ <f(x)<0

Encontrar a imagem para o intervalo −5<x<0: 425 ≤ f(x)<∞

Encontrar a imagem para o intervalo 0<x<5: −∞ <f(x)<−150

Encontrar a imagem para o intervalo 5<x<∞ : −150 <f(x)<0

Unir as imagens de todos os intervalos de domínio para obter a imagem de função

f(x)<0 or f(x)≥ 425 or f(x)<−150 or −150 <f(x)<0

f(x)<0 or f(x)≥ 425

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domain (z−8)²z²−64

range (z−8)²z²−64

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

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expandir (x^2+5)(x^2-5)

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expandir (x^2+5)(x^2-5)

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