What is the extreme f(x)=e^{6x^2+4y^2+10} ?

The extreme f(x)=e^{6x^2+4y^2+10} is Minimum(0,0)

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extreme f(x)=e^{6x^2+4y^2+10}

extreme points

f (x) = e^{6 x^{2} + 4 y^{2} + 10}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 (8 e^{6 x^{2} + 4 y^{2} + 10} y^{2} + e^{6 x^{2} + 4 y^{2} + 10})

D (x, y) = 96 (12 e^{6 x^{2} + 4 y^{2} + 10} x^{2} + e^{6 x^{2} + 4 y^{2} + 10}) (8 e^{6 x^{2} + 4 y^{2} + 10} y^{2} + e^{6 x^{2} + 4 y^{2} + 10}) - 9216 e^{2 (6 x^{2} + 4 y^{2} + 10)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

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

f (x) = - 45 x^{2} + 400 x

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

f (x) = \frac{( 2 x ^{2} + 4 x )}{( 2 + x ^{2} )}

What is the extreme f(x)=e^{6x^2+4y^2+10} ?
The extreme f(x)=e^{6x^2+4y^2+10} is Minimum(0,0)