What is the extreme f(x,y)=4-x^2-y^2 ?

The extreme f(x,y)=4-x^2-y^2 is Maximum(0,0)

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extreme f(x,y)=4-x^2-y^2

extreme points

f (x, y) = 4 - 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)

f (x) = x^{2} + 4 x

f (x, y) = 25 \cdot (y + x^{2})^{2} + (1 + x)^{2}

f (x) = sin^{2} (x) - sin (x)

f (x) = 4 x^{3} + 9 x^{2} - 12 x + 7

h (v, t) = v t - 16 t^{2}

What is the extreme f(x,y)=4-x^2-y^2 ?
The extreme f(x,y)=4-x^2-y^2 is Maximum(0,0)