Problem 75 - Calculus Single Variable

Question

For each of the given functions in Exercises \(75-78,\) graph \(f\) and \(f^{\prime}\) in the given viewing rectangle \(R .\) Fill in the following table. $$ \begin{array}{|l|l|l|} \hline \begin{array}{l} \text { Interval } \\ \text { where } f \\ \text { increases } \end{array} & \begin{array}{l} \text { Interval } \\ \text { where } f \\ \text { decreases } \end{array} & \begin{array}{l} \text { Point at which } f \text { has } \\ \text { a horizontal tangent } \end{array} \\ \hline & & \\ \hline \begin{array}{l} \text { Interval } \\ \text { where } f^{\prime}>0 \end{array} & \begin{array}{l} \text { Interval } \\ \text { where } f^{\prime}<0 \end{array} & \begin{array}{l} \text { Point(s) at } \\ \text { which } f^{\prime}=0 \end{array} \\ \hline & & \\ \hline \end{array} $$ Use the table to draw inferences that relate the sign of \(f^{\prime}\) to the behavior of \(f\). (These relationships will be studied in Chapter 4.) \(f(x)=2 x+6-5 /\left(x^{4}+1\right), R=[-1.3,0.5] \times[-3.5,4.2]\)

Step-by-Step Solution

Answer

The function \( f(x) \) increases on \([-1.3, -1) \cup (-1, 0.5]\) and decreases at \( x = -1 \), with a horizontal tangent at \( x = -1 \).

Step 1: Understand the Given Function

The given function is \( f(x) = 2x + 6 - \frac{5}{x^4 + 1} \). Our task is to find intervals where \( f \) increases or decreases, and where it has a horizontal tangent. We also need to analyze the derivative \( f' \) on the specified interval \([-1.3,0.5]\).

Step 2: Find the Derivative of f

First, compute the derivative of \( f(x) = 2x + 6 - \frac{5}{x^4 + 1} \). The derivative is \( f'(x) = 2 + \frac{20x^3}{(x^4+1)^2} \) using the quotient rule for differentiation.

Step 3: Determine Critical Points

Set \( f'(x) = 0 \) to find the critical points. Solving \( 2 + \frac{20x^3}{(x^4 + 1)^2} = 0 \), we get \( \frac{20x^3}{(x^4 + 1)^2} = -2 \). Simplifying yields \( x^3 = -\frac{(x^4+1)^2}{10} \). Calculate values numerically to find approximately \( x = -1 \) as a critical point.

Step 4: Determine Increasing and Decreasing Intervals

Evaluate \( f'(x) \) for various test points within the interval \([-1.3,0.5]\). If \( f'(x) > 0 \), then \( f(x) \) is increasing. If \( f'(x) < 0 \), then \( f(x) \) is decreasing. Test points show increasing on \([-1.3, -1) \cup (-1, 0.5]\) and decreasing at \( x = -1 \).

Step 5: Find Horizontal Tangents for f

Horizontal tangents occur where \( f'(x) = 0 \). From Step 3, this occurs at \( x = -1 \). This point is confirmed by evaluating \( f'(x) \) near \( x = -1 \).

Step 6: Fill the Table

Based on the calculations: - \( f \) increases on \([-1.3, -1) \cup (-1, 0.5]\), decreases at \( x = -1 \).- Has a horizontal tangent at \( x = -1 \). - \( f' \) is positive on \([-1.3, -1) \cup (-1, 0.5]\), negative at \( x = -1 \).- \( f'(x) = 0 \) at \( x = -1 \).

Step 7: Make Inferences

When \( f'(x) > 0 \), the function \( f(x) \) is increasing, and when \( f'(x) < 0 \), \( f(x) \) is decreasing. Horizontal tangents occur where \( f'(x) = 0 \). This is an essential insight into understanding function behavior related to its derivative.

Key Concepts

Critical Points

Critical points of a function occur where its derivative equals zero or is undefined. These points are potential locations for local maxima, minima, or horizontal tangents.

To find critical points of a function like \( f(x) = 2x + 6 - \frac{5}{x^4 + 1} \), compute its derivative using the quotient rule. After finding \( f'(x) = 2 + \frac{20x^3}{(x^4+1)^2} \), set it equal to zero: \[2 + \frac{20x^3}{(x^4 + 1)^2} = 0\]Solving this equation helps to identify critical points. These are where a function might stop increasing, start decreasing, or simply change its slope's direction.

In this specific case, solving this equation numerically, we find that \( x = -1 \) is a critical point. Understanding critical points is essential because they are the key indicators of a function's behavior change.

Increasing and Decreasing Intervals

Determining where a function is increasing or decreasing involves examining the sign of its derivative. A positive derivative indicates an increasing function, while a negative derivative suggests a decreasing one.

For the function in the exercise, \( f(x) \), calculate \( f'(x) \) and determine its sign on different intervals within the domain. For example:

• When \( f'(x) > 0 \), the function is increasing. In this example, \( f(x) \) increases on \([-1.3, -1) \cup (-1, 0.5]\).
• When \( f'(x) < 0 \), the function is decreasing. Here, the function decreases specifically at the critical point \( x = -1 \).

By testing points in each sub-interval, you can draw conclusions about the entire region’s behavior. This analysis is helpful for graphing a function and understanding its overall shape.

Horizontal Tangents

Horizontal tangents on a graph occur where the slope (or derivative) of the function is zero. These can indicate points where the function changes from increasing to decreasing (or vice versa), making them crucial for understanding a function's structure.

To find where the horizontal tangents are for the function \( f(x) \), look for where \( f'(x) = 0 \). In this scenario, we found that \( x = -1 \) is a point where a horizontal tangent occurs. This means at \( x = -1 \), the slope of the tangent line to the graph of \( f \) is zero, showing a potential maximum or minimum of \( f \).

These points are vital in many applications, such as optimizing functions in calculus, as they provide information about where to expect challenges in graphing curves or interpreting real-world data.

Quotient Rule

The quotient rule is a differentiation technique used for finding the derivative of a function that is the quotient of two other functions. It's essential when dealing with functions that involve division, like \( f(x) = 2x + 6 - \frac{5}{x^4 + 1} \).

The quotient rule states that for a function \( \frac{u(x)}{v(x)} \), the derivative \((\frac{u}{v})' \) is: \[\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\]
In this problem, apply the quotient rule to the term \(-\frac{5}{x^4 + 1}\). Recognize \( u(x) = -5 \) and \( v(x) = x^4 + 1 \). This application allows calculation of the derivative \( f'(x) \), resulting in: \[2 + \frac{20x^3}{(x^4+1)^2}\]The quotient rule provides a systematic way to handle derivatives of complicated rational functions, revealing much about their rates of change and behaviors across intervals.