What is the extreme f(x)=e^{4x^2+y^2+3} ?

The extreme f(x)=e^{4x^2+y^2+3} is Minimum(0,0)

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extreme f(x)=e^{4x^2+y^2+3}

extreme points

f (x) = e^{4 x^{2} + y^{2} + 3}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (2 e^{4 x^{2} + y^{2} + 3} y^{2} + e^{4 x^{2} + y^{2} + 3})

D (x, y) = 16 (8 e^{4 x^{2} + y^{2} + 3} x^{2} + e^{4 x^{2} + y^{2} + 3}) (2 e^{4 x^{2} + y^{2} + 3} y^{2} + e^{4 x^{2} + y^{2} + 3}) - 256 e^{2 (4 x^{2} + y^{2} + 3)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

\frac{x}{x ^{2} - 64}

f (x) = 2 ln (x) + ln (y) - 4 x - y

f (x) = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 8

f (x) = - 3 x^{2} + 7 x - y^{2} + 4 y + 7

What is the extreme f(x)=e^{4x^2+y^2+3} ?
The extreme f(x)=e^{4x^2+y^2+3} is Minimum(0,0)