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

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

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

extreme points

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

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

y = 3 sin (x) + 3 cos (x)

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

8 x^{4} - 48 x^{2}

c

f (x) = 5 + (8 + 7 x)^{\frac{2}{7}}

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