What is the extreme y=1-x^2-z^2 ?

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

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extreme y=1-x^2-z^2

extreme points

y = 1 - x^{2} - z^{2}

Maximum (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) :

Maximum

Maximum (0, 0)

20 x^{2} - 200 x + 3

y = \frac{x ^{4}}{1 + x ^{3}}

\frac{200}{x + 4 + 2 x}

f (x) = - 0.3 t^{2} + 2.4 t + 98.2

f (x) = \frac{( x ^{2} - 8 )}{( x - 3 )}