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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\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

(1, 1) :

Saddle

Check critical point

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

Maximum

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

x^{3} - 12 x^{2} - 45 x + 8

g (t) = e^{4 t^{3}} x ln (3 t - 1)

f (x) = - 4 x^{3} + 6 x^{2} + 1, - 4 \leq x \leq 3

f (x) = x^{3} - 12 x^{2} + 45 x + 9

f (x, y) = x^{- 1} - 64 y^{- 1} + x y

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