What is the extreme f(x,y)=-x^4+8x^2-y^2-1 ?

The extreme f(x,y)=-x^4+8x^2-y^2-1 is Saddle(0,0),Maximum(-2,0),Maximum(2,0)

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^4+8x^2-y^2-1

extreme points

f (x, y) = - x^{4} + 8 x^{2} - y^{2} - 1

Saddle (0, 0), Maximum (- 2, 0), Maximum (2, 0)

Solution steps

Find the critical points:

(0, 0), (- 2, 0), (2, 0)

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

D (x, y) = - 2 (- 12 x^{2} + 16)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Check critical point

(2, 0) :

Maximum

Saddle (0, 0), Maximum (- 2, 0), Maximum (2, 0)

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

\frac{x + 5}{2 - x}

f (x) = x^{2} - 8

f (x) = \frac{- 2}{x ^{2} - 9}

f (x) = - x^{4} + 3 x^{3} + 18

What is the extreme f(x,y)=-x^4+8x^2-y^2-1 ?
The extreme f(x,y)=-x^4+8x^2-y^2-1 is Saddle(0,0),Maximum(-2,0),Maximum(2,0)