What is the extreme f(x,y)=x^3+y^2-6xy-9x+8y+2 ?

The extreme f(x,y)=x^3+y^2-6xy-9x+8y+2 is Minimum(5,11),Saddle(1,-1)

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extreme f(x,y)=x^3+y^2-6xy-9x+8y+2

extreme points

f (x, y) = x^{3} + y^{2} - 6 x y - 9 x + 8 y + 2

Minimum (5, 11), Saddle (1, - 1)

Solution steps

Find the critical points:

(5, 11), (1, - 1)

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

D (x, y) = 12 x - 36

Check the sign of at each critical point

Check critical point

(5, 11) :

Minimum

Check critical point

(1, - 1) :

Saddle

Minimum (5, 11), Saddle (1, - 1)

y = sin (x) + cos (x)

f (x) = 4 x^{2} + 40, - 1 \leq x \leq 1

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

y = x 64 - x^{2}

f (x) = \frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2

What is the extreme f(x,y)=x^3+y^2-6xy-9x+8y+2 ?
The extreme f(x,y)=x^3+y^2-6xy-9x+8y+2 is Minimum(5,11),Saddle(1,-1)