What is the extreme x^3-16xy+4y^2+16x^2-12x+5 ?

The extreme x^3-16xy+4y^2+16x^2-12x+5 is Minimum(2,4),Saddle(-2,-4)

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extreme x^3-16xy+4y^2+16x^2-12x+5

extreme points

x^{3} - 16 x y + 4 y^{2} + 16 x^{2} - 12 x + 5

Minimum (2, 4), Saddle (- 2, - 4)

Solution steps

Find the critical points:

(2, 4), (- 2, - 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 48 x

Check the sign of at each critical point

Check critical point

(2, 4) :

Minimum

Check critical point

(- 2, - 4) :

Saddle

Minimum (2, 4), Saddle (- 2, - 4)

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

y = x^{3} - 3 x^{2} + 4 x - 4

f (x y) = x y

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 18

f (x) = x^{3} - 3 x^{2} - 9 x + 1, 0 \leq x \leq 5

What is the extreme x^3-16xy+4y^2+16x^2-12x+5 ?
The extreme x^3-16xy+4y^2+16x^2-12x+5 is Minimum(2,4),Saddle(-2,-4)