What is the extreme f(x,y)=3y^3+7xy-9x^2-5 ?

The extreme f(x,y)=3y^3+7xy-9x^2-5 is Saddle(0,0),Maximum(-343/2916 ,-49/162)

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extreme f(x,y)=3y^3+7xy-9x^2-5

extreme points

f (x, y) = 3 y^{3} + 7 x y - 9 x^{2} - 5

Saddle (0, 0), Maximum (- \frac{343}{2916}, - \frac{49}{162})

Solution steps

Find the critical points:

(0, 0), (- \frac{343}{2916}, - \frac{49}{162})

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 y

D (x, y) = - 324 y - 49

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{343}{2916}, - \frac{49}{162}) :

Maximum

Saddle (0, 0), Maximum (- \frac{343}{2916}, - \frac{49}{162})

f (x) = - \frac{cos ( 2 x )}{2} - 2 sin (x)

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

f (x) = e^{x^{2} - 4 x - 1}

y = e^{x} + I n x - x^{2} + \frac{3}{x ^{5}} + 3

f (x) = ln (1 + x^{2})

What is the extreme f(x,y)=3y^3+7xy-9x^2-5 ?
The extreme f(x,y)=3y^3+7xy-9x^2-5 is Saddle(0,0),Maximum(-343/2916 ,-49/162)