What is the extreme f(x)=x^3+y^2-6x^2+y-1 ?

The extreme f(x)=x^3+y^2-6x^2+y-1 is Saddle(0,-1/2),Minimum(4,-1/2)

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extreme f(x)=x^3+y^2-6x^2+y-1

extreme points

f (x) = x^{3} + y^{2} - 6 x^{2} + y - 1

Saddle (0, - \frac{1}{2}), Minimum (4, - \frac{1}{2})

Solution steps

Find the critical points:

(0, - \frac{1}{2}), (4, - \frac{1}{2})

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

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

Check the sign of at each critical point

Check critical point

(0, - \frac{1}{2}) :

Saddle

Check critical point

(4, - \frac{1}{2}) :

Minimum

Saddle (0, - \frac{1}{2}), Minimum (4, - \frac{1}{2})

- x

f (x) = \frac{3 x}{x + 2}

f (x) = 6 x + 4 y + 2 z = 24

f (x, y) = \frac{16}{x} + \frac{6}{y} + x^{2} - 3 y^{2}

f (x, y) = x^{3} - 27 x y + 27 y^{3}

What is the extreme f(x)=x^3+y^2-6x^2+y-1 ?
The extreme f(x)=x^3+y^2-6x^2+y-1 is Saddle(0,-1/2),Minimum(4,-1/2)