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

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

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

extreme points

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

Saddle (- \frac{3}{2}, 0)

Solution steps

Find the critical points:

(- \frac{3}{2}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 1

D (x, y) = - 2

Check the sign of at each critical point

Check critical point

(- \frac{3}{2}, 0) :

Saddle

Saddle (- \frac{3}{2}, 0)

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