What is the extreme f(x,y)=1-9x^2-y^2 ?

The extreme f(x,y)=1-9x^2-y^2 is Maximum(0,0)

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extreme f(x,y)=1-9x^2-y^2

extreme points

f (x, y) = 1 - 9 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) = 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = x + cos (2 x)

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, 0 \leq x \leq 10

f (x, y) = x^{2} - y^{2} - 2 x - 4 y - 4

f (x, y) = 4 x^{2} - 3 x y

f (x) = x 7 - x

What is the extreme f(x,y)=1-9x^2-y^2 ?
The extreme f(x,y)=1-9x^2-y^2 is Maximum(0,0)