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

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

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

extreme points

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

Maximum (- 1, 1)

Solution steps

Find the critical points:

(- 1, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Maximum

Maximum (- 1, 1)

f (x) = cos (x) - 6 x, 0 \leq x \leq 4 π

4 x^{3} - 12 x

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

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

f (x) = x - (- 4) x^{- 1}

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