What is the extreme f(x,y)=x^3+2y^2-27x-8y-4 ?

The extreme f(x,y)=x^3+2y^2-27x-8y-4 is Saddle(-3,2),Minimum(3,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=x^3+2y^2-27x-8y-4

extreme points

f (x, y) = x^{3} + 2 y^{2} - 27 x - 8 y - 4

Saddle (- 3, 2), Minimum (3, 2)

Solution steps

Find the critical points:

(- 3, 2), (3, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 24 x

Check the sign of at each critical point

Check critical point

(- 3, 2) :

Saddle

Check critical point

(3, 2) :

Minimum

Saddle (- 3, 2), Minimum (3, 2)

f (x) = x^{\frac{1}{3}} + 9

f (x) = (x^{2} + y^{2})^{2} = 2 \cdot (x^{2} - y^{2})

x^{3} + 3 x y^{2} - 3 x

f (x) = x^{\frac{2}{3}} (x - 2), - 2 \leq x \leq 2

f (x) = x^{2} - 2 x - lo g_{3} (y)

What is the extreme f(x,y)=x^3+2y^2-27x-8y-4 ?
The extreme f(x,y)=x^3+2y^2-27x-8y-4 is Saddle(-3,2),Minimum(3,2)