What is the extreme f(x,y)=3x^2-2x^3+y^2-8y+4 ?

The extreme f(x,y)=3x^2-2x^3+y^2-8y+4 is Minimum(0,4),Saddle(1,4)

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extreme f(x,y)=3x^2-2x^3+y^2-8y+4

extreme points

f (x, y) = 3 x^{2} - 2 x^{3} + y^{2} - 8 y + 4

Minimum (0, 4), Saddle (1, 4)

Solution steps

Find the critical points:

(0, 4), (1, 4)

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

D (x, y) = 2 (- 12 x + 6)

Check the sign of at each critical point

Check critical point

(0, 4) :

Minimum

Check critical point

(1, 4) :

Saddle

Minimum (0, 4), Saddle (1, 4)

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

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

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

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

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

What is the extreme f(x,y)=3x^2-2x^3+y^2-8y+4 ?
The extreme f(x,y)=3x^2-2x^3+y^2-8y+4 is Minimum(0,4),Saddle(1,4)