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

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inversa sin^2(x)

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

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punti estremi 3x²y+y³−3x²−3y²+2

Solution

Minimo(0, 2), Massimo(0, 0), Sella(1, 1), Sella(−1, 1)

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Definizione del Test delle Derivate Parziali Seconde

Supponiamo che (x, y)=(a, b) sia un punto critico di f(x, y)
e D(x, y)=∂²f∂ x² ∂²f∂ y² −(∂²f∂ x∂ y )². Quindi,
Se D(a, b) > 0 e ∂²f∂ x² <0 allora il punto (a, b) è un massimo locale.
Se D(a, b) > 0 e ∂²f∂ x² >0 allora il punto (a, b) è un minimo locale.
Se D(a, b)<0 allora il punto (a, b) è un punto sellare.
Se D(a, b)=0 allora il test è fallito e il punto (a, b) può essere un massimo locale, un minimo locale, un punto sellare, o nessuno.

Trova i punti critici: (0, 2), (0, 0), (1, 1), (−1, 1)

∂²f∂ x² =6y−6

D(x, y)=(6y−6)²−36x²

Controlla il segno di in ogni punto critico

Verifica il punto critico (0, 2): Minimo

Verifica il punto critico (0, 0): Massimo

Verifica il punto critico (1, 1): Sella

Verifica il punto critico (−1, 1): Sella

Minimo(0, 2), Massimo(0, 0), Sella(1, 1), Sella(−1, 1)

Solve instead

punti critici 3x²y+y³−3x²−3y²+2

Matrix QR Decomposition Examples

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