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parity x^2+(y+3)^2=16

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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critical points 3x²y+y³−3x²−3y²+2

Solution

(0, 2), (0, 0), (1, 1), (−1, 1)

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3x²y+y³−3x²−3y²+2

تعريف نقطة قوى محتملة

Critical points are points where the function is defined and its gradient is zero vector or undefined

مساوٍ لصفر f ′(x) وجد أين f(x) من أجل إيجاد النقاط القصوى المحتملة، اشتقّ

(0, 2), (0, 0), (1, 1), (−1, 1)

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extreme points 3x²y+y³−3x²−3y²+2

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

parity x^2+(y+3)^2=16

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