What is the extreme f(x,y)=4x^2+y^2+1 ?

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

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extreme f(x,y)=4x^2+y^2+1

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = 5 e^{x^{2} - 10 x}

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

- 3 x + ln (x)

f (x) = 52 (70 - x)

f (x) = x^{11} - 9 x^{7} + 5 x - 3

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