What is the extreme f(x,y)=x^2+xy+3x+2y+5 ?

The extreme f(x,y)=x^2+xy+3x+2y+5 is Saddle(-2,1)

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extreme f(x,y)=x^2+xy+3x+2y+5

extreme points

f (x, y) = x^{2} + x y + 3 x + 2 y + 5

Saddle (- 2, 1)

Solution steps

Find the critical points:

(- 2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Saddle

Saddle (- 2, 1)

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

P (a, b) = a^{3} - b^{3} + a^{2} + ab + b^{2}

f (x) = 8 x^{3} + 81 x^{2} - 42 x - 8

x^{\frac{5}{3}} - 5 x^{\frac{2}{3}}

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

What is the extreme f(x,y)=x^2+xy+3x+2y+5 ?
The extreme f(x,y)=x^2+xy+3x+2y+5 is Saddle(-2,1)