What is the extreme f(x,y)=(2x^2-3y^2)/(y^2+1) ?

The extreme f(x,y)=(2x^2-3y^2)/(y^2+1) is Saddle(0,0)

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extreme f(x,y)=(2x^2-3y^2)/(y^2+1)

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = \frac{- 16 x ^{2} y ^{2} - 16 x ^{2} + 72 y ^{2} - 24}{( y ^{2} + 1 ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

x^{3} + x, - 1 \leq x \leq 2

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

f (x) = 5 - 3 x^{2}, - 5 \leq x \leq 1

f (x) = 16 x + 21 x^{\frac{16}{21}}

e^{- y} (x^{2} + y^{2})

What is the extreme f(x,y)=(2x^2-3y^2)/(y^2+1) ?
The extreme f(x,y)=(2x^2-3y^2)/(y^2+1) is Saddle(0,0)