What is the extreme y=-sqrt(6-z^2-x^2) ?

The extreme y=-sqrt(6-z^2-x^2) is Minimum(0,0)

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extreme y=-sqrt(6-z^2-x^2)

extreme points

y = - 6 - z^{2} - x^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial z ^{2}} = \frac{6 - z ^{2}}{( 6 - z ^{2} - x ^{2} ) 6 - z ^{2} - x ^{2}}

D (z, x) = \frac{6}{( - z ^{2} - x ^{2} + 6 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x) = - \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 50

f (x) = x^{5} - 20 x^{4} + 8, 0 \leq x \leq 17

P (q, r) = 4 - q + r

\frac{d}{d x} (I n 1 - 2 x)

What is the extreme y=-sqrt(6-z^2-x^2) ?
The extreme y=-sqrt(6-z^2-x^2) is Minimum(0,0)