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

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amplitude of cos(2x+5)

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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

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points extrêmes 3x²y+y³−3x²−3y²+2

Solution

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

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Solution steps

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Définition du test de dérivée partielle seconde

En supposant que (x, y)=(a, b) est un point critique de f(x, y)
et D(x, y)=∂²f∂ x² ∂²f∂ y² −(∂²f∂ x∂ y )². Alors,
Si D(a, b) > 0 et ∂²f∂ x² <0 talors le point (a, b) est un maxima local .
Si D(a, b) > 0 et ∂²f∂ x² >0 alors le point (a, b) est un minima local.
Si D(a, b)<0 alors le point (a, b) est un point-selle.
Si D(a, b)=0 alors le test a échoué et le point (a, b) peut être un maxima local, un minima local, un point-selle ou ni l'un ni l'autre.

Trouver les points critiques: (0, 2), (0, 0), (1, 1), (−1, 1)

∂²f∂ x² =6y−6

D(x, y)=(6y−6)²−36x²

Vérifier le signe de à chaque point critique

Vérifier le point critique (0, 2): Minimum

Vérifier le point critique (0, 0): Maximum

Vérifier le point critique (1, 1): Selle

Vérifier le point critique (−1, 1): Selle

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

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points critiques 3x²y+y³−3x²−3y²+2

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