What is the extreme f(x)=e^{7x^2+y^2+1} ?

The extreme f(x)=e^{7x^2+y^2+1} is Minimum(0,0)

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extreme f(x)=e^{7x^2+y^2+1}

extreme points

f (x) = e^{7 x^{2} + y^{2} + 1}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (2 e^{7 x^{2} + y^{2} + 1} y^{2} + e^{7 x^{2} + y^{2} + 1})

D (x, y) = 28 (14 e^{7 x^{2} + y^{2} + 1} x^{2} + e^{7 x^{2} + y^{2} + 1}) (2 e^{7 x^{2} + y^{2} + 1} y^{2} + e^{7 x^{2} + y^{2} + 1}) - 784 e^{2 (7 x^{2} + y^{2} + 1)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

P (X, y) = 36 X^{2} - 9 y^{2}

\frac{5}{- 5 e x ^{2} - 5 y ^{2} + 5}

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

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

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

What is the extreme f(x)=e^{7x^2+y^2+1} ?
The extreme f(x)=e^{7x^2+y^2+1} is Minimum(0,0)