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

The extreme f(x,y)=(-2y)/(4x^2+y^2+2) is Minimum(0,sqrt(2)),Maximum(0,-sqrt(2))

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extreme f(x,y)=(-2y)/(4x^2+y^2+2)

extreme points

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

Minimum (0, 2), Maximum (0, - 2)

Solution steps

Find the critical points:

(0, 2), (0, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{4 y ( 12 x ^{2} + 6 - y ^{2} )}{( 4 x ^{2} + y ^{2} + 2 ) ^{3}}

D (x, y) = \frac{- 4096 x ^{6} - 3072 x ^{4} y ^{2} - 4096 x ^{4} - 768 x ^{2} y ^{4} - 1024 x ^{2} - 64 y ^{6} + 256 y ^{4} + 768 y ^{2}}{( y ^{2} + 2 + 4 x ^{2} ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, - 2) :

Maximum

Minimum (0, 2), Maximum (0, - 2)

f (x) = x e^{- 6 x}, 0 \leq x \leq 2

f (x) = x^{6} (x - 4)^{5}

P (q, s) = q + 8 + s

\frac{d}{d r} (rs e^{2 + r})

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