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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

3/(sqrt(5)-\sqrt{2)}-1/(sqrt(2)-1)-4/(sqrt(5)-1)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

π

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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Generated by AI

extreme points 3x²y+y³−3x²−3y²+2

Solution

Minimum(0, 2), Maximum(0, 0), Saddle(1, 1), Saddle(−1, 1)

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Solution steps

One step at a time

Second Partial Derivative Test definition

Suppose that (x, y)=(a, b) is a critical point of f(x, y)
and D(x, y)=∂²f∂ x² ∂²f∂ y² −(∂²f∂ x∂ y )². Then,
If D(a, b) > 0 and ∂²f∂ x² <0 then the point (a, b) is a local maximum.
If D(a, b) > 0 and ∂²f∂ x² >0 then the point (a, b) is a local minimum.
If D(a, b)<0 then the point (a, b) is a saddle point.
If D(a, b)=0 then test failed and the point (a, b) can be a local maximum, local minimum, saddle or neither.

Find the critical points: (0, 2), (0, 0), (1, 1), (−1, 1)

∂²f∂ x² =6y−6

D(x, y)=(6y−6)²−36x²

Check the sign of at each critical point

Check critical point (0, 2): Minimum

Check critical point (0, 0): Maximum

Check critical point (1, 1): Saddle

Check critical point (−1, 1): Saddle

Minimum(0, 2), Maximum(0, 0), Saddle(1, 1), Saddle(−1, 1)

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critical points 3x²y+y³−3x²−3y²+2

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