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

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

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

extreme points

f (x, y) = 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})

f (x) = x^{2} - 5 x + 1

f (x, y) = e^{4 x y} + 5 y

f (x) = \frac{x}{2} - cos (x)

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

f (x) = (2 cos (θ) + cos^{2} (θ))

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