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

The extreme f(x,y)=x^3-3x+y^2-4y is Saddle(-1,2),Minimum(1,2)

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

extreme points

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

Saddle (- 1, 2), Minimum (1, 2)

Solution steps

Find the critical points:

(- 1, 2), (1, 2)

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(1, 2) :

Minimum

Saddle (- 1, 2), Minimum (1, 2)

0.0025 x^{4} + 0.5 x^{3} - 8 x^{2} + 1000

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

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

f (x) = - x^{4} - 17 x^{3} - 106 x^{2} - 290 x - 293

f (x) = sin (4 x), - \frac{π}{6} \leq x \leq \frac{π}{4}

What is the extreme f(x,y)=x^3-3x+y^2-4y ?
The extreme f(x,y)=x^3-3x+y^2-4y is Saddle(-1,2),Minimum(1,2)