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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4 (2 e^{x^{2} + y^{2}} x^{2} + e^{x^{2} + y^{2}}) (2 e^{x^{2} + y^{2}} y^{2} + e^{x^{2} + y^{2}}) - 16 e^{2 (x^{2} + y^{2})} 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 y + x^{2} - y + 1

f (x) = e^{- θ ∣ x ∣}

\frac{x - 3}{x ^{2} - 18 x + 81}

x - \frac{54}{x}

f (x) = 2 x^{3} + 3 x^{2} - 120 x + 7

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