What is the extreme y=x^2-z^2 ?

The extreme y=x^2-z^2 is Saddle(0,0)

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extreme y=x^2-z^2

extreme points

y = x^{2} - z^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, z) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

ln (x) + \frac{1}{x}

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

f (x, y) = 4 x^{2} + 4 y^{2} + x^{4} + y^{4} - 6 x^{2} y^{2}

f (x, y) = 16 x y - x^{3} - 8 y^{2}