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

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

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

extreme points

f (x) = 8 y^{3} - 12 x y + x^{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) = 5 x 16 - x^{2}

3 x + 7 x^{- 1}

\frac{x ^{3} + 1}{x ^{2}}

x^{3} - 9 x^{2} + 24 x

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

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