Answer to Question #265849 in Calculus for chaitu

Question

Find the extrema of

3x^2-y^2+x^3

Answer 1

$$f(x, y)=3x^2-y^2+x^3$$

$$f_x=6x+3x^2$$

$$f_y=-2y$$

Find the critical point(s)

$$\begin{matrix} f_x=0 \\ f_y=0 \end{matrix}=>\begin{matrix} 6x+3x^2=0 \\ -2y=0 \end{matrix}$$

$Point(-2,0), Point(0,0)$

$$f_{xx}=6+6x$$

$$f_{xy}=0$$

$$f_{yy}=-2$$

$$D=\begin{vmatrix} 6+6x & 0 \\ 0 & -2 \end{vmatrix}=-12-12x$$

$Point(-2,0)$

$$f_{xx}=6+6(-2)=-6<0$$

$$D=-12-12(-2)=12>0$$

Then $f(-2, 0)$ is a local maximum.

$Point(0,0)$

$$f_{xx}=6+6(0)=6>0$$

$$D=-12-12(0)=-12<0$$

Then $Point(0,0)$ is a saddle point.

$$f(-2, 0)=3(-2)^2-(0)^2+(-2)^3=4$$