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

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

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

extreme points

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

Saddle (- \frac{3}{2}, - \frac{1}{2})

Solution steps

Find the critical points:

(- \frac{3}{2}, - \frac{1}{2})

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

D (x, y) = - \frac{4}{y ^{4}}

Check the sign of at each critical point

Check critical point

(- \frac{3}{2}, - \frac{1}{2}) :

Saddle

Saddle (- \frac{3}{2}, - \frac{1}{2})

f (x, y) = 11 x y + 5 x^{2}

ln (x^{2} - 36)

\frac{x + 2}{x - 2}

f (x) = 2 e^{- x^{2}}

f (x) = \frac{x y}{x - y}

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