f(x)=(x^4)^5

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3· (1+2)²÷ 9−12(1+(3−2)³)²

Solution

−45

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Solution steps

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(

1	2
3	4

)

Matrix Eigenwerte

Lass A eine Quadratmatrix sein, η einen Vektor und λ ein Skalar, das die Bedingung Aη=λη, erfüllt.
Dann ist λ der Eigenwert, der dem Eigenvektor λ von A zugordnet ist.

Die Eigenwerte von A sind die Wurzeln der charakteristischen Gleichung det(A−λ I)=0

det((

1	2
3	4

)−λ(

1	0
0	1

)): λ²−5λ−2

Löse λ²−5λ−2=0: λ=5+√332 , λ=5−√332

Die Eigenwerte sind:

λ=5+√332 , λ=5−√332

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

f(x)=(x^4)^5

Solution

Solution steps

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