Problem 31 - Calculus

Question

Given the numerical values shown, find approximate values for the derivative of \(f(x)\) at each of the \(x\) -values given. Where is the rate of change of \(f(x)\) positive? Where is it negative? Where does the rate of change of \(f(x)\) seem to be greatest? $$\begin{array}{l|lllllllll} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline f(x) & 18 & 13 & 10 & 9 & 9 & 11 & 15 & 21 & 30 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Derivative is positive from \( x=4 \) to \( x=7 \). It is negative from \( x=0 \) to \( x=2 \). Greatest rate of change is at \( x=7 \).

Step 1: Understand the Problem

The problem requires us to find the approximate derivative values (rate of change) of the function \( f(x) \) at each given \( x \)-value using the numerical data provided. We will use these derivatives to determine where the function's rate of change is positive (increasing), negative (decreasing), and where it is greatest.

Step 2: Calculate Derivatives at Each x-value

To find the derivative \( f'(x) \) at each \( x \)-value, we use the difference quotient formula:\[ f'(x) \approx \frac{f(x+h) - f(x)}{h} \]where \( h \) is the difference between consecutive \( x \)-values, which is 1 in this dataset. Calculate for each \( x \):- For \( x=0 \), \( f'(0) \approx \frac{f(1) - f(0)}{1} = \frac{13 - 18}{1} = -5 \)- For \( x=1 \), \( f'(1) \approx \frac{f(2) - f(1)}{1} = \frac{10 - 13}{1} = -3 \)- For \( x=2 \), \( f'(2) \approx \frac{f(3) - f(2)}{1} = \frac{9 - 10}{1} = -1 \)- For \( x=3 \), \( f'(3) \approx \frac{f(4) - f(3)}{1} = \frac{9 - 9}{1} = 0 \)- For \( x=4 \), \( f'(4) \approx \frac{f(5) - f(4)}{1} = \frac{11 - 9}{1} = 2 \)- For \( x=5 \), \( f'(5) \approx \frac{f(6) - f(5)}{1} = \frac{15 - 11}{1} = 4 \)- For \( x=6 \), \( f'(6) \approx \frac{f(7) - f(6)}{1} = \frac{21 - 15}{1} = 6 \)- For \( x=7 \), \( f'(7) \approx \frac{f(8) - f(7)}{1} = \frac{30 - 21}{1} = 9 \)

Step 3: Determine Positive and Negative Derivatives

Based on the calculated derivative values:- The derivative is positive at \( x=4 \), \( x=5 \), \( x=6 \), and \( x=7 \), indicating \( f(x) \) is increasing.- The derivative is negative at \( x=0 \), \( x=1 \), and \( x=2 \), indicating \( f(x) \) is decreasing.

Step 4: Identify Greatest Rate of Change

Among the positive derivative values, the greatest rate of change is at \( x=7 \), where \( f'(7) = 9 \). This indicates the most rapid increase in \( f(x) \) happens between \( x=7 \) and \( x=8 \).

Key Concepts

Rate of Change

In mathematics, the concept of rate of change refers to how a quantity changes over time or over another variable. It is often used to gauge how one quantity varies in relation to another. When you look at a function like \( f(x) \), the rate of change can give you insights into how \( f \) behaves as \( x \) changes.

The rate of change is positive when the function is increasing. Here, a positive rate indicates that as \( x \) increases, so does \( f(x) \). Conversely, when the rate of change is negative, the function is decreasing. This means that an increase in \( x \) leads to a decrease in \( f(x) \). Identifying these intervals can help you understand the behavior of the function and predict future values.

• A positive rate of change signifies rising trends in the graph of the function.
• A negative rate of change suggests a downward trend.

It is crucial to locate where the rate of change reaches its peak value in terms of magnitude. This maximum rate of change indicates where the function changes the most noticeably and rapidly.

Difference Quotient

The difference quotient is a fundamental concept in calculus used to approximate the derivative of a function at a certain point. The basic formula for the difference quotient is given by:\[ f'(x) \approx \frac{f(x+h) - f(x)}{h} \]Here, \( h \) denotes a small increment in \( x \), typically the distance between two consecutive \( x \)-values. It essentially measures the average rate of change of the function over the interval \([x, x+h]\). In practical terms, it serves as an approximation of how steep the tangent is at a particular point on the curve.

Calculating the difference quotient for each \( x \)-value helps us estimate the derivative, \( f'(x) \), at those points. This is useful when exact derivatives are difficult to determine analytically, or data is provided in discrete form, such as a table. From the examples given, it's evident that calculating the difference quotient for each adjacent pair of \( x \)-values allows us to estimate the behavior of \( f(x) \) effectively.

• The difference quotient simplifies the process of finding rates of change numerically.
• In this scenario, the increment \( h \) happens to be 1, simplifying calculations considerably.

Understanding and applying the difference quotient is key to transitions towards more advanced calculus concepts like solving for exact derivatives.

Numerical Calculations

Numerical calculations involve solving mathematical problems with numbers through approximation methods rather than exact analytical solutions. In the context of derivatives, numerical calculations with difference quotients allow us to work with sets of numbers, especially when data is obtained experimentally or provided in tables. They provide a practical method for estimating derivatives.

For the given exercise, these calculations involved determining derivatives numerically through predefined values of \( x \) and \( f(x) \) using a formulaic approach. Despite using an approximation, the process yields useful insights into the behavior of the function:

• It offers a detailed approximation of \( f(x) \)'s derivative without calculus.
• Handling numerical data prepares students for real-world applications.

Understanding numerical calculations is essential. Many real-world problems can't be solved analytically and require numerical methods. This approach is not just limited to derivatives but extends to many areas, including integration and differential equations.