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

The extreme f(x,y)=x^2+xy+2x+4y+1 is Saddle(-4,6)

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extreme f(x,y)=x^2+xy+2x+4y+1

extreme points

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

Saddle (- 4, 6)

Solution steps

Find the critical points:

(- 4, 6)

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

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(- 4, 6) :

Saddle

Saddle (- 4, 6)

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

f (x) = 2 x^{3} - 21 x^{2} + 60 x, 1 \leq x \leq 6

y^{3} + 6 x^{2} y - 6 x^{2} - 6 y^{2} + 3

T (x, y) = 100 + x (y + 1)^{2} - 6 x^{2} - 8 (y + 1)^{2}

\frac{( x ^{3} )}{( x - 1 ) ^{2}}

What is the extreme f(x,y)=x^2+xy+2x+4y+1 ?
The extreme f(x,y)=x^2+xy+2x+4y+1 is Saddle(-4,6)