Need This Done In Maplesoft Due Sunday At 11:59 Pm

Need This Done In Maplesoft Due Sunday At 1159pm1 Given The Funct

Given the function f(x) = x^3 − 2x + 1 and the points c₁ = -1 and c₂ = 0, perform the following tasks:

a. Calculate f(c₁). Then, determine the slopes of secant lines connecting the point (x, f(x)) to (c₁, f(c₁)), using the provided list of x-values: [-1.5, -1.3, -1.1, -0.9, -0.7, -0.5].

b. Calculate f(c₂). Next, determine the slopes of secant lines connecting (x, f(x)) to (c₂, f(c₂)), with the x-values: [-0.5, -0.3, -0.1, 0.1, 0.3, 0.5].

c. Estimate the slope of the tangent to the function at points c₁ and c₂. Provide a brief description of your observations regarding these slopes.

d. Graph the function f(x), along with the lines representing the estimated tangent lines at (c₁, f(c₁)) and (c₂, f(c₂)).

Given another function f(x) = x^3 − 2x + 4 over the interval [-1.5, 1.5], perform the following:

a. Divide the interval into 6 subintervals, find the midpoints u₁, u₂, ..., u₆ of these subintervals, then compute f(u₁), ..., f(u₆). Use these to estimate the area under the curve (via midpoint Riemann sum).

b. Partition the same interval into 3 subintervals, find their centers, compute f at these points, and again estimate the area under the curve.

c. Graph the function over the interval to compare the area estimates obtained from 6 and 3 subintervals. Do not display all rectangles, only the function graph and relevant annotations.

Paper For Above instruction

The given set of mathematical tasks involves both analytical and graphical analysis of functions, primarily focusing on derivatives, secant and tangent lines, and area approximations using Riemann sums. The analysis begins with the function f(x) = x^3 − 2x + 1, with particular points at c₁ = -1 and c₂ = 0, which serve as focal points for calculating function values and slopes. Subsequently, graphing these functions and their tangent lines provides visual insights into the behavior of the functions within specified intervals.

Part 1: The first task involves calculating f(c₁) and f(c₂). Substituting c₁ = -1 into the function yields f(-1) = (-1)^3 − 2(-1) + 1 = -1 + 2 + 1 = 2. For c₂ = 0, f(0) = 0^3 − 20 + 1 = 1. The secant slopes are then estimated by calculating (f(x) - f(c₁)) / (x - c₁) for each x in the list [-1.5, -1.3, -1.1, -0.9, -0.7, -0.5], which provides approximate derivatives around c₁. Similarly, for c₂, the same process applies with the x-values [-0.5, -0.3, -0.1, 0.1, 0.3, 0.5].

Estimating the tangent slopes at c₁ and c₂ involves taking the limit of the secant slopes as x approaches c₁ and c₂ respectively. Practically, this is approximated by selecting the secant lines with x-values closest to these points. For c₁ = -1, the slopes from points near -1 suggest the tangent slope is approximately the derivative at that point; for the function f(x), the derivative f'(x) = 3x^2 − 2, which at x = -1 gives f'(-1) = 3(1) − 2 = 1. At x = 0, f'(0) = 0 − 2 = -2, thus estimating the tangent slopes accordingly.

The visualization in the graph reveals the original function along with tangent lines at the specified points, demonstrating how the tangent line slopes compare to the secant slopes. These graphical plots serve as a valuable tool for understanding local linearity and the behavior of the function near these points.

Part 2: The second analysis involves approximating the area under the curve of f(x) = x^3 − 2x + 4 on the interval [-1.5, 1.5]. By dividing the interval into subintervals, the midpoints are used to compute the function value at u₁, u₂, ..., u₆ for the 6-part partition, then summing these with the width of subintervals to approximate the area via the midpoint Riemann sum. Similarly, dividing into only 3 subintervals and calculating the midpoints provides an alternative, cruder approximation.

This comparative approach illustrates how the granularity of partitioning impacts the accuracy of the area estimation. Graphing these subdivisions alongside the function curve helps visualize the rectangles' placement and the approximation's precision. A narrower subdivision (more subintervals) results in a more accurate estimate.

Graphical representation for both cases involves plotting the continuous function and overlaying the rectangles representing the Riemann sum rectangles, excluding details of all rectangles per the instructions. This visual comparison emphasizes the convergence of the midpoint sum values as the number of subintervals increases.

In summary, the tasks integrate concepts of differential calculus, including secant and tangent lines, with integral approximation techniques such as Riemann sums, complemented by graphical visualization to enhance understanding of the functions' behaviors and the accuracy of numerical methods.

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