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.

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