What is the extreme f(x)=2xy-15y^2 ?

The extreme f(x)=2xy-15y^2 is Saddle(0,0)

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extreme f(x)=2xy-15y^2

extreme points

f (x) = 2 x y - 15 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 30

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 6 sin (x) + cot (x)

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

f (x, y) = 5 + 4 x + 6 y - x^{2} - 3 y^{2}

f (x) = 48 π r^{2} + \frac{1804.8}{r}

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