What is the extreme f(x,y)=-x^2+8-y^2 ?

The extreme f(x,y)=-x^2+8-y^2 is Maximum(0,0)

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extreme f(x,y)=-x^2+8-y^2

extreme points

f (x, y) = - x^{2} + 8 - 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)

f (x) = 2 x^{2} + \frac{28}{x}

f (x) = ln (x^{2} + 3 x + 7), - 2 \leq x \leq 2

g (x) = x^{3} - x^{2} - x + 3

f (x) = - x^{2} + 10 x - 21, 3 \leq x \leq 7

f (x) = 4 x^{5} - 25 x^{4} - 40 x^{3} - 4

What is the extreme f(x,y)=-x^2+8-y^2 ?
The extreme f(x,y)=-x^2+8-y^2 is Maximum(0,0)