Problem 13 - Calculus Early Transcendentals

Question

A function \(y=f(x)\) and values of \(x_{0}\) and \(x_{1}\) are given. (a) Find the average rate of change of \(y\) with respect to \(x\) over the interval \(\left[x_{0}, x_{1}\right] .\) (b) Find the instantaneous rate of change of \(y\) with respect to \(x\) at the specified value of \(x_{0}\) (c) Find the instantaneous rate of change of \(y\) with respect to \(x\) a) The arbitrary value of \(x_{0} .\) (d) The average rate of change in part (a) is the slope of a certain secant line, and the instantaneous rate of change in part (b) is the slope of a certain tangent line. Sketch the graph of \(y=f(x)\) together with those two lines. $$ y=1 / x ; x_{0}=2, x_{1}=3 $$

Step-by-Step Solution

Answer

(a) -1/6; (b) -1/4; (c) -1/x_0^2.

Step 1: Identify the Function and Interval

We are given the function \(y = \frac{1}{x}\) and need to calculate over the interval \([x_0, x_1] = [2, 3]\).

Step 2: Calculate the Average Rate of Change

The average rate of change of a function over an interval \([x_0, x_1]\) is given by \(\frac{f(x_1) - f(x_0)}{x_1 - x_0}\). For \(y = \frac{1}{x}\), this becomes: \[ \frac{f(3) - f(2)}{3 - 2} = \frac{\frac{1}{3} - \frac{1}{2}}{1} = \frac{-\frac{1}{6}}{1} = -\frac{1}{6}. \]

Step 3: Find the Instantaneous Rate of Change at \(x_0 = 2\)

The instantaneous rate of change is equivalent to the derivative of the function evaluated at that point. The derivative of \(y = \frac{1}{x}\) is \(f'(x) = -\frac{1}{x^2}\). Evaluating this at \(x = 2\) gives: \[ f'(2) = -\frac{1}{2^2} = -\frac{1}{4}. \]

Step 4: Find the Instantaneous Rate of Change at Arbitrary \(x_0\)

Using the derivative \(f'(x) = -\frac{1}{x^2}\), the instantaneous rate of change at any point \(x_0\) is: \[ f'(x_0) = -\frac{1}{x_0^2}. \]

Step 5: Sketch the Graph and Lines

On a coordinate plane, plot the graph of \(y = \frac{1}{x}\). Draw a secant line connecting the points \((2, f(2))\) and \((3, f(3))\), where the average slope calculated was \(-\frac{1}{6}\). Also draw a tangent line at \(x_0 = 2\) with the slope \(-\frac{1}{4}\), representing the instantaneous rate of change at this point.

Key Concepts

Average Rate of Change

The average rate of change of a function provides us with a way to measure how much the function's value changes, on average, over a certain interval. It's essentially the "overall slope" of the function between two points. To find it, we use the formula:

• \( \frac{f(x_1) - f(x_0)}{x_1 - x_0} \)

This formula tells us how the values of the function at two points, \(x_0\) and \(x_1\), affect the function's change over that interval.
In the example with the function \(y = \frac{1}{x}\), the average rate of change from \(x=2\) to \(x=3\) is \(-\frac{1}{6}\). This means that, on average, for every unit \(x\) increases within this interval, \(y\) decreases by \( \frac{1}{6}\).

Instantaneous Rate of Change

The instantaneous rate of change of a function at a certain point gives us the rate at which the function's value is changing exactly at that point, almost like a "snapshot" of its behavior. This is calculated using the derivative of the function. For example, at \(x_0 = 2\) for the function \(y = \frac{1}{x}\), the instantaneous rate of change is \(-\frac{1}{4}\). This tells us how steep the graph is at that particular point, showing the immediate change occurring at \(x=2\).
Derivatives are powerful because they help us understand how a function behaves locally, rather than over an interval.

Derivative

The derivative of a function is a fundamental concept in calculus, representing the function's instantaneous rate of change. Mathematically, the derivative is defined as:

• \( f'(x) \)

For the function \(y = \frac{1}{x}\), its derivative is \( f'(x) = -\frac{1}{x^2} \).
This derivative tells us the slope of the tangent line to the graph of \(y\) at any point \(x\). It's a powerful tool for analyzing function behavior, allowing us to explore how \(y\) changes as \(x\) changes, without having to consider large intervals.

Secant Line

A secant line is a straight line that passes through two points on the graph of a function. The slope of the secant line represents the average rate of change of the function between these two points.
In the given exercise, the secant line between \( (2, f(2)) \) and \( (3, f(3)) \) on the graph of \(y = \frac{1}{x}\) shows the average rate of change to be \(-\frac{1}{6}\).

• This line helps visualize and compare how the function behaves between theses points and is essential in understanding overall trends on an interval.

Tangent Line

The tangent line to a function at a given point gives us the slope of the function at that exact spot, and it doesn't touch the curve at any other point close by (within a small vicinity). It's the graphical representation of the instantaneous rate of change.

• In the current scenario at \(x_0 = 2\) for \(y = \frac{1}{x}\), the slope of this line is \(-\frac{1}{4}\).

This line is crucial for understanding how sharply a function is changing right at a point, providing key insights into its local behavior.
Tangent lines are used in approximating values and solving problems where a precise understanding of function behavior at a point is needed.