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

The extreme f(x,y)=1-(x^2+y^2) is Maximum(0,0)

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

\frac{x}{( x ^{2} + 1 )}

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

f (x) = 4 x^{5} - 7 x^{2} + 8

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

f (x) = 5 x^{3} + x^{4}

What is the extreme f(x,y)=1-(x^2+y^2) ?
The extreme f(x,y)=1-(x^2+y^2) is Maximum(0,0)