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

The extreme f(x,y)=e^{x^2+y^2-2x} is Minimum(1,0)

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

extreme points

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

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

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

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

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

4 x^{3} - 12 x^{2}

3 x y - x^{2} y - x y^{2}

f (x) = x^{3} - 4 x^{2} + 4 x - 1

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x - y + 10

f (x) = \frac{x ^{4}}{4} - 8 x

What is the extreme f(x,y)=e^{x^2+y^2-2x} ?
The extreme f(x,y)=e^{x^2+y^2-2x} is Minimum(1,0)