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domain of inverse f(x)=x^2

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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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Steps Examples

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eigenvektoren (

1	2	3
4	5	6
7	8	9

)

Solution

(

−2

), (

−2(11−3√33)

11+3√33

), (

−2(3√33+11)

−3√33+11

)

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Solution steps

One step at a time

(

1	2	3
4	5	6
7	8	9

)

Matrix Eigenvektoren

Lass A eine Quadratmatrix, η ein Vektor und λ ein Skalar sein, welches diese Bedingung erfüllt: Aη=λη,
dann ist η ein Eigenvektor von A und λ ist ein Eigenwert, der damit verbunden ist.

Um die Eigenvektoren η zu finden, löse (A−λ I)η=0 für jeden Eigenwert λ

Ermittle die Eigenwerte von (

1	2	3
4	5	6
7	8	9

): λ=0, λ=15+3√332 , λ=15−3√332

Eigenvektoren für λ=0: (

−2

)

Eigenvektoren für λ=15+3√332 : (

−2(11−3√33)

11+3√33

)

Eigenvektoren für λ=15−3√332 : (

−2(3√33+11)

−3√33+11

)

Die Eigenvektoren für (

1	2	3
4	5	6
7	8	9

)

−2

), (

−2(11−3√33)

11+3√33

), (

−2(3√33+11)

−3√33+11

)

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1	2	3
4	5	6
7	8	9

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

domain of inverse f(x)=x^2

Solution

Solution steps

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