Problem 50 - University Calculus: Early Transcendentals

Question

Use a CAS to perform the following steps for the functions. a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$f(x)=x+\frac{5}{x}, \quad x_{0}=1$$

Step-by-Step Solution

Answer

The limit of the difference quotient as \( h \to 0 \) is 4, and it represents the slope of the tangent line at \( x = 1 \).

Step 1: Plot the Function

First, we need to define the function \( f(x) = x + \frac{5}{x} \). Next, we plot this function over the interval \(ig(1 - \frac{1}{2}\big) \leq x \leq (1 + 3)\) or \(\frac{1}{2} \leq x \leq 4\). Use your CAS (Computer Algebra System) tool to input the function and generate this plot, highlighting the behavior of \( f(x) \) within this range.

Step 2: Define the Difference Quotient

The difference quotient is given by \( q(h) = \frac{f(1+h) - f(1)}{h} \). Substitute \( f(x) = x + \frac{5}{x} \) into this expression. Calculate \( f(1+h) = (1+h) + \frac{5}{1+h} \) and \( f(1) = 1 + 5 = 6 \). Now, the difference quotient becomes \( q(h) = \frac{(1+h) + \frac{5}{1+h} - 6}{h} \). Enter this into your CAS to simplify and work with it further.

Step 3: Calculate the Limit as h Approaches Zero

We find the limit of \( q(h) \) as \( h \rightarrow 0 \). Using your CAS, compute \( \lim_{{h \to 0}} \frac{(1+h) + \frac{5}{1+h} - 6}{h} \). Simplify the expression step-by-step to get the derivative at \( x_0 = 1 \), which gives the slope of the tangent line. The limit resolves to 4.

Step 4: Define and Graph Secant Lines

Define the equation of secant lines using \( y = f(1) + q(h) \cdot (x - 1) \) for different values of \( h \) such as 3, 2, and 1. Use each \( q(h) \), and compute values using your CAS. Then, graph these secant lines along with the function \( f(x) \) and the tangent line (from the limit found in Step 3). This should be plotted over the interval \([0.5, 4]\). Use the derivative at \( x_0 = 1 \) to define the tangent line: \( y = 6 + 4(x - 1) \).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in differential calculus. It is used to approximate the derivative of a function at a particular point. The formula for the difference quotient is given by:

• \[ q(h) = \frac{f(x_0 + h) - f(x_0)}{h} \]

Here, \( h \) represents a small change in \( x \), and \( x_0 \) is the point at which we want to find the derivative. In simpler terms, the difference quotient measures how much the function \( f \) changes for a small change \( h \) in \( x \).
For the function \( f(x) = x + \frac{5}{x} \) at the point \( x_0 = 1 \), we compute \( f(1+h) = (1+h) + \frac{5}{1+h} \). Substituting \( f(1) = 6 \) and reorganizing gives the difference quotient:

• \[ q(h) = \frac{(1+h) + \frac{5}{1+h} - 6}{h} \]

This expression becomes vital when we calculate the derivative by taking the limit as \( h \to 0 \).

Secant Line

Secant lines are lines that intersect a curve at two or more points. They represent an average rate of change of the function across a specified interval.
In the context of our function, secant lines help us visualize how the curve of \( f(x) = x + \frac{5}{x} \) changes across small distances. For different values of \( h \), such as 3, 2, and 1, the secant line is given by:

• \[ y = f(1) + q(h) \cdot (x - 1) \]

Here, the slope \( q(h) \) varies with the step size \( h \), showing us different approximations to the true tangent line at \( x = 1 \).
By graphing these lines, we observe how secant lines provide a bridge to the concept of the tangent line, as \( h \) becomes very small.

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. It represents the instantaneous rate of change of the function at that point.
For the function \( f(x) = x + \frac{5}{x} \) at \( x_0 = 1 \), the tangent line is found by determining the derivative, or the slope of the function at that point. From the limit we computed earlier, the slope is 4.
Thus, the equation of the tangent line is:

• \[ y = 6 + 4(x - 1) \]

This line provides an approximation of the curve near \( x = 1 \) and is tangent to the curve at precisely this point. It shows the function's behavior very locally.

Limit

In calculus, the concept of limits is essential in understanding how functions behave as inputs approach certain values. When it comes to the difference quotient and derivative calculations, limits play a critical role.
The goal is to evaluate the limit of the difference quotient \( q(h) \) as \( h \to 0 \). By finding \[ \lim_{{h \to 0}} \frac{ (1+h) + \frac{5}{1+h} - 6}{h} \], we simplify the expression to find the exact slope of the tangent line.
This limit process essentially bridges the gap between secant and tangent lines by showing how the average rate of change transforms into the instantaneous rate of change. For our function, this limit results in 4, which is the derivative of \( f \) at \( x_0 = 1 \). The concept of limits allows us to understand and compute derivatives, which are fundamental in calculus for analyzing and interpreting functions.