What is the extreme f(x)=xy^2-y^2+1/2 x^2-5x+4 ?

The extreme f(x)=xy^2-y^2+1/2 x^2-5x+4 is Minimum(5,0),Saddle(1,2),Saddle(1,-2)

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extreme f(x)=xy^2-y^2+1/2 x^2-5x+4

extreme points

f (x) = x y^{2} - y^{2} + \frac{1}{2} x^{2} - 5 x + 4

Minimum (5, 0), Saddle (1, 2), Saddle (1, - 2)

Solution steps

Find the critical points:

(5, 0), (1, 2), (1, - 2)

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

D (x, y) = 2 x - 2 - 4 y^{2}

Check the sign of at each critical point

Check critical point

(5, 0) :

Minimum

Check critical point

(1, 2) :

Saddle

Check critical point

(1, - 2) :

Saddle

Minimum (5, 0), Saddle (1, 2), Saddle (1, - 2)

0.001 x^{2} + 7 + \frac{128}{x}

f (t) = \frac{1}{9} [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 28 t]

F (X, Y) = X \cdot Y

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

What is the extreme f(x)=xy^2-y^2+1/2 x^2-5x+4 ?
The extreme f(x)=xy^2-y^2+1/2 x^2-5x+4 is Minimum(5,0),Saddle(1,2),Saddle(1,-2)