What is the extreme x^2+x-3xy+y^3-5 ?

The extreme x^2+x-3xy+y^3-5 is Minimum(1,1),Saddle(1/4 , 1/2)

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extreme x^2+x-3xy+y^3-5

extreme points

x^{2} + x - 3 x y + y^{3} - 5

Minimum (1, 1), Saddle (\frac{1}{4}, \frac{1}{2})

Solution steps

Find the critical points:

(1, 1), (\frac{1}{4}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 12 y - 9

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(\frac{1}{4}, \frac{1}{2}) :

Saddle

Minimum (1, 1), Saddle (\frac{1}{4}, \frac{1}{2})

y = x e^{3 x^{2}}

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

f (x, y) = 4 x + 6 y - x^{2} - y^{2} + 6

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

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

What is the extreme x^2+x-3xy+y^3-5 ?
The extreme x^2+x-3xy+y^3-5 is Minimum(1,1),Saddle(1/4 , 1/2)