answer:To find the equation of the plane, we first need to determine a normal vector to the plane. We can find this by taking the cross product of two vectors lying on the plane. We can form two vectors from the given points: overrightarrow{P_1P_2} and overrightarrow{P_1P_3}. Let overrightarrow{P_1P_2} = color{blue}{(-2,-3,4) - (-2,3,-1)} = color{blue}{(0,-6,5)} and overrightarrow{P_1P_3} = color{blue}{(0,-3,1) - (-2,3,-1)} = color{blue}{(2,-6,2)}. Now, calculate the cross product: overrightarrow{N} = overrightarrow{P_1P_2} times overrightarrow{P_1P_3} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 0 & -6 & 5 2 & -6 & 2 end{vmatrix} = (-6 cdot 2 - (-6) cdot 5) mathbf{i} - (0 cdot 2 - (-6) cdot 2) mathbf{j} + (0 cdot (-6) - (-6) cdot 2) mathbf{k} overrightarrow{N} = (12 + 30) mathbf{i} - (0 - 12) mathbf{j} + (0 + 12) mathbf{k} = color{blue}{42 mathbf{i} + 12 mathbf{j} + 12 mathbf{k}} The equation of the plane is given by: (mathbf{r} - mathbf{r_0}) cdot overrightarrow{N} = 0 where mathbf{r} is any point on the plane, and mathbf{r_0} is a known point (we'll use P_1). Let mathbf{r} = color{blue}{(x, y, z)} and mathbf{r_0} = color{blue}{(-2, 3, -1)}, so (mathbf{r} - mathbf{r_0}) cdot overrightarrow{N} = (color{blue}{(x, y, z)} - color{blue}{(-2, 3, -1)}) cdot color{blue}{(42, 12, 12)} Rightarrow (x + 2, y - 3, z + 1) cdot (42, 12, 12) = 0 Rightarrow 42(x + 2) + 12(y - 3) + 12(z + 1) = 0 Rightarrow 42x + 84 + 12y - 36 + 12z + 12 = 0 Rightarrow 42x + 12y + 12z + 60 = 0 Dividing by 12 for simplicity: Rightarrow 3.5x + y + z + 5 = 0 Rightarrow 3.5x + y + z = -5 The equation of the plane is 3.5x + y + z - 5 = 0.

answer:This sequence is a special case of a more general sequence studied by Alman, Cuenca, and Huang in their paper "Laurentness of Somos-4 Sequences." In their Theorem 3.9.(6), they prove that the terms of this sequence are always integers. Their proof is based on the theory of caterpillar sequences developed by Fomin and Zelevinsky, and later refined by Lam and Pylyavskyy. The key idea is to show that the sequence can be expressed as a Laurent series in terms of five formal variables x_1, x_2, x_3, x_4, x_5. That is, each term of the sequence can be written as a quotient of two polynomials in these variables, where the denominator is a monomial. Alman, Cuenca, and Huang provide an algorithm to test whether a given polynomial is "1-periodic," which means that it can be used to generate a Laurent series with integer coefficients. They show that the polynomial defining the given sequence is 1-periodic, which implies that the terms of the sequence are always integers.

answer:The Jacobian matrix of a vector-valued function mathbf{r}(x, y, z) = langle f(x, y, z), g(x, y, z), h(x, y, z) rangle is a matrix whose entries are the partial derivatives of the component functions f, g, and h. In this case, we have: f(x, y, z) = e^x g(x, y, z) = tan(y) h(x, y, z) = frac{z}{x} So, the Jacobian matrix of mathbf{r}(x, y, z) is: J(mathbf{r}(x, y, z)) = begin{bmatrix} frac{partial}{partial x} [e^x] & frac{partial}{partial y} [e^x] & frac{partial}{partial z} [e^x] frac{partial}{partial x} [tan(y)] & frac{partial}{partial y} [tan(y)] & frac{partial}{partial z} [tan(y)] frac{partial}{partial x} [frac{z}{x}] & frac{partial}{partial y} [frac{z}{x}] & frac{partial}{partial z} [frac{z}{x}] end{bmatrix} = begin{bmatrix} e^x & 0 & 0 0 & sec^2(y) & 0 -frac{z}{x^2} & 0 & frac{1}{x} end{bmatrix} The Jacobian matrix of mathbf{r}(x, y, z) is given by: J(mathbf{r}(x, y, z)) = begin{bmatrix} frac{partial}{partial x} [e^x] & frac{partial}{partial y} [e^x] & frac{partial}{partial z} [e^x] frac{partial}{partial x} [tan(y)] & frac{partial}{partial y} [tan(y)] & frac{partial}{partial z} [tan(y)] frac{partial}{partial x} [frac{z}{x}] & frac{partial}{partial y} [frac{z}{x}] & frac{partial}{partial z} [frac{z}{x}] end{bmatrix} = begin{bmatrix} e^x & 0 & 0 0 & sec^2(y) & 0 -frac{z}{x^2} & 0 & frac{1}{x} end{bmatrix}

answer:Since the function is an open downward parabola, its maximum value is at the vertex. We can find the x-coordinate of the vertex by setting the derivative equal to zero: {eq}begin{align*} 6 - 2x &= 0 2x &= 6 x &= 3 end{align*} {/eq} The maximum value of the function is {eq}f(3) = 5 {/eq}. Since the parabola opens downward, the range of the function is {eq}(-infty, 5] {/eq}.