What is the extreme f(x)= 1/(sqrt(16-x^2-y^2)) ?

The extreme f(x)= 1/(sqrt(16-x^2-y^2)) is Minimum(0,0)

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extreme f(x)= 1/(sqrt(16-x^2-y^2))

extreme points

f (x) = \frac{1}{16 - x ^{2} - y ^{2}}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 y ^{2} - x ^{2} + 16}{( 16 - x ^{2} - y ^{2} ) ^{\frac{5}{2}}}

D (x, y) = \frac{- 2 x ^{4} - 4 x ^{2} y ^{2} + 16 x ^{2} - 2 y ^{4} + 16 y ^{2} + 256}{( - x ^{2} - y ^{2} + 16 ) ^{5}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

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

f (x, y) = - 4 y^{2} - 5 x^{3} + 13 x^{2}

f (x, y) = x - \frac{1}{2} y^{2} - \frac{1}{3} x^{3} + y + 6

f (x) = \frac{4 x ^{2}}{2 x - 8}

What is the extreme f(x)= 1/(sqrt(16-x^2-y^2)) ?
The extreme f(x)= 1/(sqrt(16-x^2-y^2)) is Minimum(0,0)