What is the extreme f(x,y)=e^{3x^2+y^2+6} ?

The extreme f(x,y)=e^{3x^2+y^2+6} is Minimum(0,0)

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extreme f(x,y)=e^{3x^2+y^2+6}

extreme points

f (x, y) = e^{3 x^{2} + y^{2} + 6}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (2 e^{3 x^{2} + y^{2} + 6} y^{2} + e^{3 x^{2} + y^{2} + 6})

D (x, y) = 12 (6 e^{3 x^{2} + y^{2} + 6} x^{2} + e^{3 x^{2} + y^{2} + 6}) (2 e^{3 x^{2} + y^{2} + 6} y^{2} + e^{3 x^{2} + y^{2} + 6}) - 144 e^{2 (3 x^{2} + y^{2} + 6)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = x^{4} - 4 x + 7

f (x) = 400 - \frac{3600}{x + 8} - x

f (x) = 3 x^{2} - 2 x^{2} - 5 x + 6

f (x) = I n (x + 1 + x^{2})

f (x) = x^{4} - 4 x - 8

What is the extreme f(x,y)=e^{3x^2+y^2+6} ?
The extreme f(x,y)=e^{3x^2+y^2+6} is Minimum(0,0)