Question

Let \(x=2\) and complete the table for the function. $$ \begin{array}{|c|c|c|c|c|} \hline d x=\Delta x & d y & \Delta y & \Delta y-d y & d y / \Delta y \\ \hline 1.000 & & & & \\ \hline 0.500 & & & & \\ \hline 0.100 & & & & \\ \hline 0.010 & & & & \\ \hline 0.001 & & & & \\ \hline \end{array} $$ \(y=\frac{1}{x}\)

Short answer

The table after computations should be filled as follows: $$\begin{array}{|c|c|c|c|c|}\hline dx=Δx & dy & Δy & Δy−dy & dy/Δy \\hline 1.000 & -0.5 & -0.333 & 0.167 & 1.5 \\hline 0.500 & -0.25 & -0.154 & 0.096 & 1.621 \\hline 0.100 & -0.05 & -0.029 & 0.021 & 1.724 \\hline 0.010 & -0.005 & -0.003 & 0.002 & 1.727 \\hline 0.001 & -0.0005 & -0.0003 & 0.0002 & 1.732 \\hline\end{array}$$

Step-by-step solution

- Compute dy

Compute \(dy\) for given \(x=2\) and \(\Delta x\). For a function \(y = 1/x\), \(dy = -1/(x^2)\cdot dx\). These are the new \(y\) values when \(x\) changes by \(dx\). Substitute \(x=2\) and each of the \(dx\) values (1, 0.5, 0.1, 0.01, 0.001) into this equation to get the corresponding \(dy\) values.

- Compute ∆y

Next, compute \(\Delta y\) by evaluating the function \(y = 1/x\) at new \(x\) values, that is, at each \(x + \Delta x\), and then subtracting the original \(y\) value (which is \(1/2\) for \(x=2\)). Now, the new \(x\) values will be \(2+\Delta x\) (where \(\Delta x\) being 1, 0.5, 0.1, 0.01, 0.001), and the corresponding \(\Delta y\) values can be calculated.

- Compute ∆y - dy

Calculate \(\Delta y- dy\) by simply subtracting the \(dy\) values calculated in step 1 from the \(\Delta y\) values calculated in step 2.

- Compute dy/∆y

Next step is to compute the ratio of \(dy\) to \(\Delta y\) using the computed \(dy\) and \(\Delta y\) values.

Key concepts

Derivatives

The concept of a derivative is at the heart of differential calculus. It is a measure of how a function's value changes as its input changes. Mathematically, a derivative is represented as the limit of the average rate of change of the function as the interval of change becomes infinitesimally small.

In the given exercise, the derivative is calculated for the function \(y = \frac{1}{x}\) at \(x = 2\). The resultant value, \(dy\), represents the instantaneous rate of change of \(y\) with respect to a small change in \(x\), labeled as \(dx\). Using the formula for the derivative of \(y\), which is \(-\frac{1}{x^2}\), we input the \(dx\) values to find how \(y\) changes at \(x = 2\). This gives us insight into the behavior of the function at a specific point and is fundamental for understanding how functions behave more generally.

Function Analysis

Function analysis involves looking at the various features of functions such as their derivatives, turning points, asymptotes, and points of inflection. Through function analysis, we seek to understand the behavior of functions over their entire domain. For instance, in analyzing \(y = \frac{1}{x}\), we observe that the function decreases as \(x\) increases, and it approaches zero but never actually reaches it.

The process of completing the table in the exercise adds a practical dimension to function analysis, as it requires computing and comparing the derivative \(dy\) with the actual change in \(y\), represented as \(\Delta y\). By making these calculations for different values of \(\Delta x\), students learn to appreciate the varying rates of change and how the theoretical derivative compares to the practical changes observed in the function's output.

Limits

Limits are a fundamental concept in calculus that describe the behavior of a function as its input approaches a particular value. It is crucial for defining derivatives and integrals. A limit poses the question: as we get closer and closer to a certain \(x\) value, what value does the function \(y\) approach?

In the context of our exercise, limits help in understanding the derivative. The calculation of \(dy\) for various increments of \(\Delta x\) is essentially exploring the limit of the function's rate of change as \(\Delta x\) approaches zero. We see that as \(\Delta x\) gets smaller and smaller (approaching zero), the value of \(dy\) corresponds more closely with that of \(\Delta y\), reflecting the concept that the derivative is the limit of the difference quotient as \(\Delta x\) goes to zero.

Rate of Change

The rate of change of a function is a measure of how quickly the function's value is changing at any given point. It's an indication of the function's behavior and can be constant or variable. For example, the rate of change for a linear function is constant, while for nonlinear functions, like our given function \(y = \frac{1}{x}\), the rate of change varies with \(x\).

The step-by-step solution of the exercise required calculating two types of change: \(d y\) which is the theoretical rate of change as predicted by calculus (the derivative), and \(\Delta y\), which is the actual change in the function's value as \(x\) is increased by a finite amount \(\Delta x\). By comparing these, students understand how derivatives approximate real changes in a function's value for small changes in \(x\), and they become more familiar with the concept of rate of change as a foundational idea in differential calculus.