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

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

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

extreme points

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

Minimum (2, 0)

Solution steps

Find the critical points:

(2, 0)

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

D (x, y) = 4 e^{x^{2} + y^{2} - 4 x} (2 x^{2} - 8 x + 9) (2 e^{x^{2} + y^{2} - 4 x} y^{2} + e^{x^{2} + y^{2} - 4 x}) - 4 e^{2 (x^{2} + y^{2} - 4 x)} y^{2} (2 x - 4)^{2}

Check the sign of at each critical point

Check critical point

(2, 0) :

Minimum

Minimum (2, 0)

f (x, y) = (x - y^{2}) (x - y^{3})

f (x) = x^{4} - 18 x^{2} + 4

f (x) = 49 x + \frac{16}{x}

f (x) = x^{2} + 10 x + 9

f (x) = x - 2, 0 \leq x \leq 4

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