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

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

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

extreme points

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

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 2

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

y = x^{4} - 2 x^{3}

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

f (x) = x^{3} - 6 x + 9 x - 2

f (x) = x^{3} - 12 x, 0 \leq x \leq 5

f (x) = 2 x^{3} - 6 x^{2} + 4

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