What is the extreme f(x,y)=x^3+2xy+y^2-5x ?

The extreme f(x,y)=x^3+2xy+y^2-5x is Minimum(5/3 ,-5/3),Saddle(-1,1)

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extreme f(x,y)=x^3+2xy+y^2-5x

extreme points

f (x, y) = x^{3} + 2 x y + y^{2} - 5 x

Minimum (\frac{5}{3}, - \frac{5}{3}), Saddle (- 1, 1)

Solution steps

Find the critical points:

(\frac{5}{3}, - \frac{5}{3}), (- 1, 1)

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

D (x, y) = 12 x - 4

Check the sign of at each critical point

Check critical point

(\frac{5}{3}, - \frac{5}{3}) :

Minimum

Check critical point

(- 1, 1) :

Saddle

Minimum (\frac{5}{3}, - \frac{5}{3}), Saddle (- 1, 1)

f (x) = - x^{3} - 7 x^{2} - 16 x - 42

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

x + \frac{64}{x}

f (x) = x^{3} - 6 x^{2} - 15 x + 10

f (x) = x^{4} - 32 x^{2} + 6

What is the extreme f(x,y)=x^3+2xy+y^2-5x ?
The extreme f(x,y)=x^3+2xy+y^2-5x is Minimum(5/3 ,-5/3),Saddle(-1,1)