answer:To find the zeros of f(x) = x^4 - 6x^3 + 11x^2 - 6x, we first factor out an x from the polynomial, yielding f(x) = x(x^3 - 6x^2 + 11x - 6). This immediately tells us that x = 0 is a zero of the function. Now, we need to find the zeros of the cubic polynomial g(x) = x^3 - 6x^2 + 11x - 6. At first glance, solving this cubic equation seems daunting, but let's try to factor it. After inspection, we notice that g(1) = 1 - 6 + 11 - 6 = 0. This means x = 1 is a zero of g(x). We can perform polynomial division to factor g(x) as (x - 1)(x^2 - 5x + 6). Next, we factor the quadratic polynomial x^2 - 5x + 6. This quadratic factors as (x - 2)(x - 3), giving us the remaining zeros of g(x) as x = 2 and x = 3. Therefore, the zeros of f(x) are x = 0, x = 1, x = 2, and x = 3. This problem showcases how factoring can be a powerful tool in finding zeros of polynomials.

answer:To determine if the function f(x) = e^(2x) / (e^x - 1) has a horizontal asymptote as x approaches infinity, we can examine the behavior of the function as x becomes large. We can rewrite f(x) as f(x) = e^x / (1 - e^(-x)). As x approaches infinity, e^(-x) approaches 0, so the denominator approaches 1. Meanwhile, the numerator, e^x, grows exponentially large. However, to better understand the behavior of f(x), we can divide both the numerator and the denominator by e^x, which is the dominant term. This manipulation gives us f(x) = e^x / (e^x - 1) = 1 / (1 - e^(-x)). Now, as x approaches infinity, e^(-x) approaches 0, and the denominator approaches 1. Therefore, f(x) approaches 1 as x approaches infinity. This implies that the function f(x) has a horizontal asymptote of y = 1 as x approaches infinity, but it approaches this asymptote from above, since f(x) > 1 for all x > 0.

answer:The sum of the geometric series 1 + r + r^2 + ... can be found by considering the series itself and a related series obtained by multiplying the original series by r. Let S be the sum of the series: S = 1 + r + r^2 + r^3 + ... Multiplying S by r yields: rS = r + r^2 + r^3 + r^4 + ... Subtracting rS from S, we get: S - rS = 1 This simplifies to: S(1 - r) = 1 Now, dividing both sides by (1 - r), we obtain: S = 1 / (1 - r) This formula holds when |r| < 1, because in that case, the series converges, meaning its sum approaches a finite value. If |r| ≥ 1, the series diverges, either growing without bound or oscillating indefinitely. The formula S = 1 / (1 - r) provides the sum of the geometric series when the absolute value of the common ratio r is less than 1.

answer:The determinant of a matrix has a rich geometric interpretation, offering insights into the nature of linear transformations and the relationships between vectors. Consider a square matrix A, representing a linear transformation that maps vectors from one space to another. The determinant of A, denoted as det(A), can be viewed as a scaling factor that describes how this transformation affects the volume or area of the space. In the context of a 2x2 matrix, the determinant represents the scaling factor by which the linear transformation changes the area of a parallelogram formed by two vectors. Specifically, if we have two vectors v and w, the area of the parallelogram they span is given by the magnitude of their cross product. When we apply the linear transformation represented by the matrix A, the new area is scaled by a factor equal to the absolute value of det(A). If det(A) is positive, the orientation of the parallelogram is preserved; if it is negative, the orientation is reversed. This idea generalizes to higher dimensions. For a 3x3 matrix, the determinant represents the scaling factor for the volume of a parallelepiped formed by three vectors. The sign of the determinant again indicates whether the orientation of the parallelepiped is preserved or reversed under the linear transformation. The geometric interpretation of the determinant also provides insight into the invertibility of a matrix. If the determinant is non-zero, the linear transformation is invertible, meaning that it is possible to map the transformed space back to the original space. Geometrically, this means that the transformation does not collapse the space onto a lower-dimensional subspace. On the other hand, a zero determinant indicates that the transformation is not invertible, and the space is collapsed, resulting in a loss of information. This connection between the determinant and the geometry of linear transformations highlights the deep relationship between algebraic and geometric concepts in mathematics, demonstrating how the tools of linear algebra can be used to describe and analyze the properties of spaces and transformations.