Question

Let \(\quad y=5 x-x^{2}\), and take \(x_{0}=1\); then \(y_{0}=4\). If \(\Delta x=.2\), then \(\Delta y=.56\) and \(\frac{\Delta y}{\Delta x}=2.8\). Show further that, and for \(\Delta x=.1, \quad \Delta y=.29, \quad \frac{\Delta y}{\Delta x}=2.9 ;\) for \(\Delta x=.01, \quad \Delta y=.0299, \quad \frac{\Delta y}{\Delta x}=2.99 .\)

Short answer

Question: Verify if the given conclusions for the function \(y = 5x - x^2\) are correct, when \(x_0 = 1\), \(y_0 = 4\), and the following \(\Delta x\), \(\Delta y\) and \(\frac{\Delta y}{\Delta x }\) values are provided: 1. \(\Delta x = 0.2\), \(\Delta y = 0.56\), and \(\frac{\Delta y}{\Delta x} = 2.8\) 2. \(\Delta x = 0.1\), \(\Delta y = 0.29\), and \(\frac{\Delta y}{\Delta x} = 2.9\) 3. \(\Delta x = 0.01\), \(\Delta y = 0.0299\), and \(\frac{\Delta y}{\Delta x} = 2.99\) Answer: Yes, the given conclusions are correct, as the calculated \(\frac{\Delta y}{\Delta x}\) values match the values provided in the exercise.

Step-by-step solution

Calculate function values for given points

Given the function \(y = 5x - x^2\), we know that when \(x_0=1\), \(y_0=4\). Let's calculate the function value for each \(\Delta x\) value provided: For \(\Delta x = 0.2\), we have \(x_1 = x_0 + \Delta x = 1.2\), and the function value at this point is: \(y_1 = 5(1.2) - (1.2)^2 = 4.56\) For \(\Delta x = 0.1\), we have \(x_2 = x_0 + \Delta x = 1.1\), and the function value at this point is: \(y_2 = 5(1.1) - (1.1)^2 = 4.29\) For \(\Delta x = 0.01\), we have \(x_3 = x_0 + \Delta x = 1.01\), and the function value at this point is: \(y_3 = 5(1.01) - (1.01)^2 = 4.0299\)

Calculate the change in \(y\), \(\Delta y\)

Now, we will calculate the change in the \(y\) values, which is denoted as \(\Delta y\). It can be calculated as the difference between the function values: For \(\Delta x = 0.2\), we have: \(\Delta y_1 = y_1 - y_0 = 4.56 - 4 = 0.56\) For \(\Delta x = 0.1\), we have: \(\Delta y_2 = y_2 - y_0 = 4.29 - 4 = 0.29\) For \(\Delta x = 0.01\), we have: \(\Delta y_3 = y_3 - y_0 = 4.0299 - 4 = 0.0299\)

Calculate the rate of change, \(\frac{\Delta y}{\Delta x}\)

With the calculated \(\Delta y\) values, we can now compute the rate of change by dividing \(\Delta y\) by \(\Delta x\). For \(\Delta x = 0.2\), we have: \(\frac{\Delta y_1}{\Delta x} = \frac{0.56}{0.2} = 2.8\) For \(\Delta x = 0.1\), we have: \(\frac{\Delta y_2}{\Delta x} = \frac{0.29}{0.1} = 2.9\) For \(\Delta x = 0.01\), we have: \(\frac{\Delta y_3}{\Delta x} = \frac{0.0299}{0.01} = 2.99\) The calculated \(\frac{\Delta y}{\Delta x}\) values match the given exercise's values, which demonstrates that the given conclusions are correct.

Key concepts

Differential Calculus

Differential calculus is fundamental when we need to understand how things change. It's a major branch of calculus that deals with the concept of a derivative, which itself represents an instantaneous rate of change of a function concerning one of its variables. In essence, differential calculus provides a systematic way to calculate and analyze the rate at which quantities change.

For instance, in the given exercise, we explore how a change in the input variable, denoted as \(\Delta x\), leads to a change in the output variable, represented by \(\Delta y\). As \(\Delta x\) gets smaller, we approximate the derivative of the function by calculating the ratio \(\frac{\Delta y}{\Delta x}\). This ratio represents the average rate of change over the interval defined by \(\Delta x\). As \(\Delta x\) approaches zero, the ratio \(\frac{\Delta y}{\Delta x}\) approaches the derivative of the function at a point. The derivative is often written as \(\frac{dy}{dx}\) or \(y'\), and it's a powerful tool in function analysis to determine slope, velocity, acceleration, and many other concepts that involve rate of change.

Function Analysis

Function analysis is an aspect of calculus that involves examining different properties of functions, such as continuity, limits, and behavior changes. Through function analysis, we can determine where a function is increasing or decreasing, identify its maxima and minima, and establish points of inflection where the curvature changes.

For the function in our exercise, \(y=5x-x^2\), function analysis would involve investigating its behavior as \(x\) changes. This particular quadratic function is a parabola, and by calculating the derivative \(y'\), we can determine its slope at any given point on the curve. By analyzing the values for \(\frac{\Delta y}{\Delta x}\) given in the exercise, we can infer that between \(x_0=1\) and higher values of \(x\), the function is increasing but at a decreasing rate. This is a crucial insight that can lead to better graph interpretation and understanding of a function's dynamics over its domain.

Incremental Change

Incremental change captures the essence of calculus—'the mathematics of small differences.' When studying the behavior of functions, we often consider how small changes in input can affect the output. The variable \(\Delta x\) signifies a small, finite increment in the function's independent variable, and \(\Delta y\) represents the corresponding change in the dependent variable.

As illustrated in the exercise's solution, calculating \(\Delta y\) for increasingly smaller values of \(\Delta x\) can help us understand how the function \(y=5x-x^2\) reacts to these changes. The exercise improvement advice suggests that demonstrating this process for successively smaller increments of \(x\) helps build an intuitive grasp of the function's behavior near \(x_0=1\). Ultimately, incremental change is the stepping stone to understanding derivatives and, by extension, differential calculus as a whole. It illustrates that calculus is not just about abstract concepts but about measuring and predicting the behavior of real-world phenomena.