평균 변화율 f(x)=5xln(x), [1, e^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

simplify solve for inverse tangent line

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eigenvectors (

1	0	0
0	1	0
0	0	1

)

Solution

(

), (

)

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Solution steps

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(

1	0	0
0	1	0
0	0	1

)

Matrix eigenvectors

Let A be a square matrix, η a vector and λ a scalar that satisfy Aη=λη,
then η is an eigenvector of A and λ is the eigenvalue associated with it

To find the eigenvectors η, solve (A−λ I)η=0 for each eigenvalue λ

Find the eigenvalues for (

1	0	0
0	1	0
0	0	1

): λ=1 with multiplicity of 3

Eigenvectors for λ=1: (

), (

)

), (

)

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1	0	0
0	1	0
0	0	1

) or choose below

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Notebook

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

평균 변화율 f(x)=5xln(x), [1, e^2]

Solution

Solution steps

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