What is the extreme f(x)=x^3-12xy+8y^3 ?

The extreme f(x)=x^3-12xy+8y^3 is Saddle(0,0),Minimum(2,1)

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extreme f(x)=x^3-12xy+8y^3

extreme points

f (x) = x^{3} - 12 x y + 8 y^{3}

Saddle (0, 0), Minimum (2, 1)

Solution steps

Find the critical points:

(0, 0), (2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 48 y

D (x, y) = 288 x y - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 1) :

Minimum

Saddle (0, 0), Minimum (2, 1)

f (x) = x 9 - x^{2}, - 1 \leq x \leq 3

x^{2} - 2 x - 8

f (x) = 2 x - y

f (x) = 2 x + y

y = \frac{x ^{2}}{x ^{2} + 3}

What is the extreme f(x)=x^3-12xy+8y^3 ?
The extreme f(x)=x^3-12xy+8y^3 is Saddle(0,0),Minimum(2,1)