answer:The given equation can be rewritten and classified as follows: Classification: Ellipse Standard Form: 6 left(x + frac{1}{12}right)^2 + 8 left(y + frac{1}{16}right)^2 = frac{487}{96} This can be simplified to the standard form of an ellipse with the following properties: Center: left(-frac{1}{12}, -frac{1}{16}right) Major Axis: Along the y-axis (since a^2 = 8 > b^2 = 6) Eccentricity (e): frac{1}{2} (as it is an ellipse) Semi-major axis (a): frac{sqrt{frac{487}{96}}}{sqrt{8}} = frac{sqrt{487}}{16sqrt{2}} Semi-minor axis (b): frac{sqrt{frac{487}{96}}}{sqrt{6}} = frac{sqrt{487}}{12sqrt{2}} Foci: Located at a distance of ae from the center along the major axis. Therefore, the foci are: left( begin{array}{cc} -frac{1}{12} - frac{sqrt{487}}{24sqrt{2}} & -frac{1}{16} -frac{1}{12} + frac{sqrt{487}}{24sqrt{2}} & -frac{1}{16} end{array} right) Area Enclosed: The area (A) of an ellipse is given by A = pi ab. Therefore, the area enclosed by this ellipse is: A = pi left(frac{sqrt{487}}{16sqrt{2}}right)left(frac{sqrt{487}}{12sqrt{2}}right) = frac{487pi}{384sqrt{3}} In summary, the equation represents an ellipse with a center at left(-frac{1}{12}, -frac{1}{16}right), eccentricity frac{1}{2}, and an area of frac{487pi}{384sqrt{3}}.

answer:To find the points of intersection, we set the two functions equal to each other: {eq}f(x) = g(x) x^3 - x^2 + x + 1 = x^3 + x^2 + x - 1 text{Simplify and solve for } x: -x^2 + 1 = x^2 - 1 2x^2 = 2 x^2 = 1 x = pm1 {eq} Now, we'll use one of the functions, say {eq}f(x){/eq}, to find the corresponding y-coordinates: For {eq}x = -1: y = (-1)^3 - (-1)^2 + (-1) + 1 = -1 - 1 - 1 + 1 = -2 text{So, one point of intersection is } P_1(-1, -2). {eq} For {eq}x = 1: y = 1^3 - 1^2 + 1 + 1 = 1 - 1 + 1 + 1 = 2 text{Thus, the second point of intersection is } P_2(1, 2). {eq} Hence, the coordinates of the points where the graphs intersect are {eq}P_1(-1, -2){/eq} and {eq}P_2(1, 2){/eq}.

answer:The Jacobian matrix of mathbf{r}(x, y, z) is given by: J(mathbf{r}(x, y, z)) = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} Calculating each partial derivative, we get: frac{partial f}{partial x} = 0, quad frac{partial f}{partial y} = sec^2(y), quad frac{partial f}{partial z} = 0 frac{partial g}{partial x} = 0, quad frac{partial g}{partial y} = -frac{1}{2y^{3/2}}, quad frac{partial g}{partial z} = 0 frac{partial h}{partial x} = cos(x + y), quad frac{partial h}{partial y} = cos(x + y), quad frac{partial h}{partial z} = 0 Therefore, the Jacobian matrix of mathbf{r}(x, y, z) is: J(mathbf{r}(x, y, z)) = begin{bmatrix} 0 & sec^2(y) & 0 0 & -frac{1}{2y^{3/2}} & 0 cos(x + y) & cos(x + y) & 0 end{bmatrix}

answer:Consider the following construction: f(x) = inf{ |x-y| : yin mathbb Rsetminus V} where V is the given non-empty open set. This function satisfies the following properties: * f(x)=0 if and only if xnotin V. * f(x)>0 if and only if xin V. * f(x) is continuous, as it satisfies the Lipschitz condition: |f(x) - f(x')| le |x-x'|. Therefore, this function meets the desired criteria.