What is the extreme f(x)=((x-1)^2+(y+2)^2)/2 ?

The extreme f(x)=((x-1)^2+(y+2)^2)/2 is Minimum(1,-2)

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extreme f(x)=((x-1)^2+(y+2)^2)/2

extreme points

f (x) = \frac{( x - 1 ) ^{2} + ( y + 2 ) ^{2}}{2}

Minimum (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 1

Check the sign of at each critical point

Check critical point

(1, - 2) :

Minimum

Minimum (1, - 2)

f (x) = - 7 (2 x^{3} - \frac{125}{32})^{2} + 1

f (x, y) = \frac{1}{( x + y )}

f (x) = \frac{1}{1 - x ^{2}}

f (x) = \frac{x ^{2}}{x + 1}, - 2 \leq x \leq 2

f (x) = \frac{- x - 8}{10 x ^{2} - 40 x + 27}

What is the extreme f(x)=((x-1)^2+(y+2)^2)/2 ?
The extreme f(x)=((x-1)^2+(y+2)^2)/2 is Minimum(1,-2)