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domain of cos(x)+sin(x)

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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

lhopital squeeze chain factoring sandwich

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Steps Examples

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domain 6a²+5a+10−(4a²+6a+12)

Solution

−∞ <a<∞

Interval Notation

(−∞ , ∞ )

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Solution steps

One step at a time

(

1+	2+
3+	4

)

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+	+2+
3+	+4

): λ=5+√332 , λ=5−√332

Eigenvectors for λ=5+√332 : (

−3+√33

)

Eigenvectors for λ=5−√332 : (

−√33−3

)

The eigenvectors for (

1+	+2+
3+	+4

)

−3+√33

), (

−√33−3

)

Solve instead

range 6a²+5a+10−(4a²+6a+12)

vertices 6a²+5a+10−(4a²+6a+12)

inverse 6a²+5a+10−(4a²+6a+12)

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

domain of cos(x)+sin(x)

Solution

Solution steps

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