What is the extreme f(x)=y^3-3xy+6x ?

The extreme f(x)=y^3-3xy+6x is Saddle(4,2)

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extreme f(x)=y^3-3xy+6x

extreme points

f (x) = y^{3} - 3 x y + 6 x

Saddle (4, 2)

Solution steps

Find the critical points:

(4, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(4, 2) :

Saddle

Saddle (4, 2)

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

f (x) = 3 sin^{2} (x) + 3 sin (x)

f (x) = x^{3} - 12 x^{2} - 45 x + 1

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

h (t, w) = \frac{1}{6} tw

What is the extreme f(x)=y^3-3xy+6x ?
The extreme f(x)=y^3-3xy+6x is Saddle(4,2)