What is the extreme f(x)=3x^2+4y^3-12xy+37 ?

The extreme f(x)=3x^2+4y^3-12xy+37 is Minimum(4,2),Saddle(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3x^2+4y^3-12xy+37

extreme points

f (x) = 3 x^{2} + 4 y^{3} - 12 x y + 37

Minimum (4, 2), Saddle (0, 0)

Solution steps

Find the critical points:

(4, 2), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y

D (x, y) = 144 y - 144

Check the sign of at each critical point

Check critical point

(4, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (4, 2), Saddle (0, 0)

f (x) = 15 x + 18 y - 2 x^{2} + 2 x y + 3 y^{2}

f (x, y) = x^{3} - 12 x + y^{2} + 6 y + 14

f (x) = x^{- 7} ln (x)

f (x, y) = y (e^{x} - 1)

y = \frac{5 x ^{2}}{( x - 4 ) ^{2}}

What is the extreme f(x)=3x^2+4y^3-12xy+37 ?
The extreme f(x)=3x^2+4y^3-12xy+37 is Minimum(4,2),Saddle(0,0)