Find and classify all critical points for the function f(x, y) = 2y³ + 3y² - 12y -x² + 2x. A) (1, -2) local maximum; (1, 1) local minimum B) (1, -2) local maximum; (1, 1) saddle point C) (1, -2) local maximum; (1, 1) local maximum D) (1, -2) local minimum; (1, 1) local minimum E) (-1, 2) saddle point; (1, 1) saddle point

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Quizplus · Accepted Answer

To find and classify the critical points of the function $f(x, y) = 2y^3 + 3y^2 - 12y - x^2 + 2x$, we first find the partial derivatives and set them equal to zero to find critical points. Then, we use the second derivative test to classify these points.1. Find the partial derivatives:$\frac{\partial f}{\partial x} = -2x + 2$,$\frac{\partial f}{\partial y} = 6y^2 + 6y - 12$.2. Set the partial derivatives equal to zero to find critical points:$-2x + 2 = 0$ gives $x = 1$,$6y^2 + 6y - 12 = 0$ simplifies to $y^2 + y - 2 = 0$, which factors to $(y + 2)(y - 1) = 0$, giving $y = -2$ and $y = 1$.So, the critical points are $(1, -2)$ and $(1, 1)$.3. To classify these points, we use the second derivative test, which involves computing the second partial derivatives and evaluating the determinant of the Hessian matrix at each critical point.The second partial derivatives are:$\frac{\partial^2 f}{\partial x^2} = -2$,$\frac{\partial^2 f}{\partial y^2} = 12y + 6$,$\frac{\partial^2 f}{\partial x \partial y} = 0$.The Hessian determinant at a point $(x, y)$ is given by $D = f_{xx}f_{yy} - (f_{xy})^2$.For $(1, -2)$, $D = (-2)(12(-2) + 6) - 0^2 = -2(-18) = 36 > 0$ and $f_{xx} = -2 < 0$, indicating a local maximum.For $(1, 1)$, $D = (-2)(12(1) + 6) - 0^2 = -2(18) = -36 < 0$, indicating a saddle point.Therefore, the correct answer is (1, -2) is a local maximum and (1, 1) is a saddle point.

Find and Classify All Critical Points for the Function F(x