thừa số x^2-5x+6

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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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critical points f(x)=1x²

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Global extreme points definition

f(x) has a global maximum at x₀ on the interval (a, b) if f(x₀) ≥ f(x) for all x in (a, b)
f(x) has a global minimum at x₀ on the interval (a, b) if f(x₀) ≤ f(x) for all x in (a, b)

Find the extreme points: Maximum(−5+2π2 +πn, 1), Minimum(3π−52 +πn, −1)

Find the function range: [−1, 1]

Verify extreme points are global points

Global Maximum(−5+2π2 +πn, 1), Global Minimum(3π−52 +πn, −1)

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domain 1x²

range 1x²

inverse 1x²

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

thừa số x^2-5x+6

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