What is the extreme f(x,y)=4x^3+2y^2-24x^2+2y+8 ?

The extreme f(x,y)=4x^3+2y^2-24x^2+2y+8 is Saddle(0,-1/2),Minimum(4,-1/2)

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extreme f(x,y)=4x^3+2y^2-24x^2+2y+8

extreme points

f (x, y) = 4 x^{3} + 2 y^{2} - 24 x^{2} + 2 y + 8

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}} = 4

D (x, y) = 4 (24 x - 48)

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})

\frac{8 x}{1 - x}

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

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

f (x, y) = x^{2} + y^{2} + x y - 8 x - 7 y + 400

f (x) = \frac{( x ^{2} - 12 )}{x - 4}

What is the extreme f(x,y)=4x^3+2y^2-24x^2+2y+8 ?
The extreme f(x,y)=4x^3+2y^2-24x^2+2y+8 is Saddle(0,-1/2),Minimum(4,-1/2)