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

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

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extreme f(x)=sqrt(1-2x^2)-y^2

extreme points

f (x) = 1 - 2 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) = \frac{4 - 2 x ^{2} + 1}{( - 2 x ^{2} + 1 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = e^{5 x} + e^{- x}

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

x^{5} - 15 x^{3}

f (x, y) = ln (4 x^{2} + 18 y^{2} - 36)

F (x, y) = x^{3} - 3 x y + 3 y^{2} + 1

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