What is the extreme f(x,y)=4x^2-12x+y^2+2y-10 ?

The extreme f(x,y)=4x^2-12x+y^2+2y-10 is Minimum(3/2 ,-1)

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extreme f(x,y)=4x^2-12x+y^2+2y-10

extreme points

f (x, y) = 4 x^{2} - 12 x + y^{2} + 2 y - 10

Minimum (\frac{3}{2}, - 1)

Solution steps

Find the critical points:

(\frac{3}{2}, - 1)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(\frac{3}{2}, - 1) :

Minimum

Minimum (\frac{3}{2}, - 1)

x^{4} - 8 x^{2} + 1

f (x, y) = \frac{9}{3} x^{2} y + y^{3} - 2 x y^{2}

f (x) = 1 + \frac{1}{x} + \frac{1}{x ^{2}}

f (x) = \frac{x + 1}{x ^{2}}

f (x, y) = 3 - 25 - (x - 2)^{2} - 4 (y + 1)^{2}

What is the extreme f(x,y)=4x^2-12x+y^2+2y-10 ?
The extreme f(x,y)=4x^2-12x+y^2+2y-10 is Minimum(3/2 ,-1)