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

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

Popular >

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( y ^{2} + 8 ) ( - 2 x ^{2} + 16 )}{( 16 - y ^{2} ) ^{\frac{5}{2}}}

D (x, y) = - \frac{8 ( y ^{2} + 8 ) ( - 2 x ^{2} + 16 )}{( 16 - y ^{2} ) ^{3}} - \frac{16 x ^{2} y ^{2}}{( 16 - y ^{2} ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 8 + (9 + 5 x)^{\frac{2}{5}}

f (x, y) = 7 x^{2} + 8 y^{2} + 4 x - 5 y + 8

f (x) = - 7 sec (x), - \frac{π}{3} \leq x \leq \frac{π}{3}

f (x, y) = \frac{x y ( 9 - x - 2 y )}{3}

x^{3} + 3 x y^{2} - 3 y^{2} - 3 x^{2} + 11

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