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

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})

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)