Let $f ( x , y ) = x ^ { 3 } + y ^ { 3 } + 3 y ^ { 2 } - 3 x - 9 y + 2$ . Then the set of saddle points of f is

A) $\{ ( 1 , - 3,27 ) \}$
B) $\{ ( - 1,1 , - 1 ) \}$
C) $\{ ( 1 , - 3,27 ) , ( - 1,1 , - 1 ) \}$
D) $\{ ( 1 , - 3,27 ) , ( - 1,1,1 ) \}$
E) $\{ ( 1 , - 3,27 ) , ( 1,1 , - 1 ) \}$

Question

Let $f ( x , y ) = x ^ { 3 } + y ^ { 3 } + 3 y ^ { 2 } - 3 x - 9 y + 2$ . Then the set of saddle points of f is

A) $\{ ( 1 , - 3,27 ) \}$
B) $\{ ( - 1,1 , - 1 ) \}$
C) $\{ ( 1 , - 3,27 ) , ( - 1,1 , - 1 ) \}$
D) $\{ ( 1 , - 3,27 ) , ( - 1,1,1 ) \}$
E) $\{ ( 1 , - 3,27 ) , ( 1,1 , - 1 ) \}$

Quizplus · Accepted Answer

The saddle points of f are the points where the first partial derivatives are both zero and the second partial derivatives have opposite signs.$f_x = 3x^2 - 3 = 0$$f_y = 3y^2 + 6y - 9 = 0$Solving these equations, we get the critical points $(1, -3)$ and $(-1, 1)$.$f_{xx} = 6x$$f_{yy} = 6y + 6$$f_{xy} = 0$At $(1, -3)$, $f_{xx} = 6 > 0$ and $f_{yy} = -12 < 0$, so this is a saddle point.At $(-1, 1)$, $f_{xx} = -6 < 0$ and $f_{yy} = 12 > 0$, so this is also a saddle point.Therefore, the set of saddle points of f is $\{ ( 1 , - 3,27 ) , ( - 1,1 , - 1 ) \}$.

Let f(x,y)=x3+y3+3y2−3x−9y+2f ( x , y ) = x ^ { 3 } + y ^ { 3 } + 3 y ^ { 2 } - 3 x - 9 y + 2f(x,y)=x3+y3+3y2−3x−9y+2

Let $f ( x , y ) = x ^ { 3 } + y ^ { 3 } + 3 y ^ { 2 } - 3 x - 9 y + 2$