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

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

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

extreme points

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

Maximum (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) :

Maximum

Maximum (0, 0)

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

f (x) = 5 x^{3} - 60 x + 5

\frac{5 x}{x ^{2} - 4}

\frac{x ^{2} + 3 x}{x - 1}

f (x) = \frac{x}{x ^{2} + 12 x + 32}

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