What is the extreme f(x)=4x^2+8x+4y^2+4 ?

The extreme f(x)=4x^2+8x+4y^2+4 is Minimum(-1,0)

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extreme f(x)=4x^2+8x+4y^2+4

extreme points

f (x) = 4 x^{2} + 8 x + 4 y^{2} + 4

Minimum (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 64

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Minimum (- 1, 0)

f (x) = 3 x^{2} + 2 x - 1

f (x) = x^{2} + 8 x + 17

f (x) = 2 x + 1

f (x, y) = 2 x + 3 y + 5

What is the extreme f(x)=4x^2+8x+4y^2+4 ?
The extreme f(x)=4x^2+8x+4y^2+4 is Minimum(-1,0)