What is the extreme f(x)= 1/(sqrt(16-x^2+4y^2)) ?

The extreme f(x)= 1/(sqrt(16-x^2+4y^2)) is Saddle(0,0)

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extreme f(x)= 1/(sqrt(16-x^2+4y^2))

extreme points

f (x) = \frac{1}{16 - x ^{2} + 4 y ^{2}}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 ( ( 16 - x ^{2} + 4 y ^{2} ) ^{\frac{3}{2}} - 12 y ^{2} 4 y ^{2} + 16 - x ^{2} )}{( 16 - x ^{2} + 4 y ^{2} ) ^{3}}

D (x, y) = \frac{- 8 ( x ^{2} + 2 ( y ^{2} + 4 ) ) ( ( 16 - x ^{2} + 4 y ^{2} ) ^{\frac{3}{2}} - 12 y ^{2} 4 y ^{2} + 16 - x ^{2} ) ( 4 y ^{2} + 16 - x ^{2} ) ^{5} - 144 x ^{2} y ^{2} ( 4 y ^{2} + 16 - x ^{2} ) ^{\frac{11}{2}}}{( 4 y ^{2} + 16 - x ^{2} ) ^{\frac{21}{2}}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{3} (1 - x)^{2}

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

f (x) = (x^{2} - 4)^{7}

y = x^{3} - 27 x + 3, - 2 \leq x \leq 4

f (x) = 2 x^{4} - 8 x + 3

What is the extreme f(x)= 1/(sqrt(16-x^2+4y^2)) ?
The extreme f(x)= 1/(sqrt(16-x^2+4y^2)) is Saddle(0,0)