Let $f ( x , y ) = 2 x ^ { 4 } + y ^ { 2 } - x ^ { 2 } - 2 y$ . Then, the set of saddle points of f is

A) $\{ ( 0,2 , - 1 ) \}$
B) $\{ ( 0,1,1 ) \}$
C) $\{ ( 0,1 , - 1 ) \}$
D) $\{ ( 0 , - 1,1 ) \}$
E) $\{ ( 0 , - 1 , - 1 ) \}$

Question

Let $f ( x , y ) = 2 x ^ { 4 } + y ^ { 2 } - x ^ { 2 } - 2 y$ . Then, the set of saddle points of f is

A) $\{ ( 0,2 , - 1 ) \}$
B) $\{ ( 0,1,1 ) \}$
C) $\{ ( 0,1 , - 1 ) \}$
D) $\{ ( 0 , - 1,1 ) \}$
E) $\{ ( 0 , - 1 , - 1 ) \}$

Quizplus · Accepted Answer

Explanation:To find the saddle points, we need to find the points where the first partial derivatives are both zero and the second partial derivatives have opposite signs.$$\frac{\partial f}{\partial x} = 8x^3 - 2x = 0 \Rightarrow x = 0, \pm \frac{1}{\sqrt{2}}$$$$\frac{\partial f}{\partial y} = 2y - 2 = 0 \Rightarrow y = 1$$So, the critical points are $(0,1), (\pm \frac{1}{\sqrt{2}}, 1)$.At $(0,1)$, the second partial derivatives are:$$\frac{\partial^2 f}{\partial x^2} = 24x^2 = 0$$$$\frac{\partial^2 f}{\partial y^2} = 2 > 0$$So, $(0,1)$ is a saddle point.At $\left(\pm \frac{1}{\sqrt{2}}, 1\right)$, the second partial derivatives are:$$\frac{\partial^2 f}{\partial x^2} = 24\left(\pm \frac{1}{\sqrt{2}}\right)^2 = 12 > 0$$$$\frac{\partial^2 f}{\partial y^2} = 2 > 0$$So, $\left(\pm \frac{1}{\sqrt{2}}, 1\right)$ are not saddle points.Therefore, the set of saddle points of f is $\{ ( 0,1 , - 1 ) \}$.

Let f(x,y)=2x4+y2−x2−2yf ( x , y ) = 2 x ^ { 4 } + y ^ { 2 } - x ^ { 2 } - 2 yf(x,y)=2x4+y2−x2−2y

Let $f ( x , y ) = 2 x ^ { 4 } + y ^ { 2 } - x ^ { 2 } - 2 y$