What is the extreme f(x)=e^{x^2+5y^2+12} ?

The extreme f(x)=e^{x^2+5y^2+12} is Minimum(0,0)

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extreme f(x)=e^{x^2+5y^2+12}

extreme points

f (x) = e^{x^{2} + 5 y^{2} + 12}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10 (10 e^{x^{2} + 5 y^{2} + 12} y^{2} + e^{x^{2} + 5 y^{2} + 12})

D (x, y) = 20 (2 e^{x^{2} + 5 y^{2} + 12} x^{2} + e^{x^{2} + 5 y^{2} + 12}) (10 e^{x^{2} + 5 y^{2} + 12} y^{2} + e^{x^{2} + 5 y^{2} + 12}) - 400 e^{2 (x^{2} + 5 y^{2} + 12)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = ln (4 - x^{2} - y^{2}) - \frac{1}{1 - x ^{2}}

y = 0.16 t e^{- 0.6 t}

f (x) = - \frac{1}{x}, - 2 \leq x \leq - 1

x^{6} - 10 x^{5} - 400 x^{4} + 2500 x^{3}

13 x^{2} - 273 x + 10840

What is the extreme f(x)=e^{x^2+5y^2+12} ?
The extreme f(x)=e^{x^2+5y^2+12} is Minimum(0,0)