Problem 39 - Calculus - AP Edition

Question

The function \(f\) whose graph is shown below has values as given in the accompanying table. $$\begin{array}{|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 1.56 & 0.58 & 2.12 & 2.34 & 2.2 \\ \hline \end{array}$$ (GRAPH CAN'T COPY) (a) Use data from the table to calculate the difference quotients $$\frac{f(3)-f(1)}{3-1}, \quad \frac{f(2)-f(1)}{2-1}, \quad \frac{f(2)-f(0)}{2-0}$$ (b) Using the graph of \(y=f(x),\) indicate which difference quotient in part (a) best approximates \(f^{\prime}(1)\) and which difference quotient gives the worst approximation to \(f^{\prime}(1)\)

Step-by-Step Solution

Answer

Best approximation: \( \frac{f(2) - f(1)}{2 - 1} \); Worst: \( \frac{f(2) - f(0)}{2 - 0} \).

Step 1: Calculate the first difference quotient

The first difference quotient is \( \frac{f(3) - f(1)}{3 - 1} \). Use the provided table to find \( f(3) = 2.2 \) and \( f(1) = 2.12 \). Substitute these values into the formula to get \[\frac{2.2 - 2.12}{3 - 1} = \frac{0.08}{2} = 0.04.\]

Step 2: Calculate the second difference quotient

The second difference quotient is \( \frac{f(2) - f(1)}{2 - 1} \). From the table, find \( f(2) = 2.34 \) and \( f(1) = 2.12 \). Plug these into the formula to get \[\frac{2.34 - 2.12}{2 - 1} = \frac{0.22}{1} = 0.22.\]

Step 3: Calculate the third difference quotient

The third difference quotient is \( \frac{f(2) - f(0)}{2 - 0} \). From the table, find \( f(2) = 2.34 \) and \( f(0) = 0.58 \). Substitute these values:\[\frac{2.34 - 0.58}{2 - 0} = \frac{1.76}{2} = 0.88.\]

Step 4: Analyze the best approximation for the derivative

The best approximation for \( f'(1) \) should ideally involve points close to \( x = 1 \). The difference quotient \( \frac{f(2) - f(1)}{2 - 1} = 0.22 \) involves points directly addressing \( x = 1 \), providing a closer approximation for \( f'(1) \).Compare this with the general behavior observed from the graph for additional context.

Step 5: Determine the worst approximation for the derivative

The worst approximation for \( f'(1) \) typically comes from using points further away from \( x = 1 \). The difference quotient \( \frac{f(2) - f(0)}{2 - 0} = 0.88 \) uses points farther away from \( x = 1 \), making it less ideal to approximate \( f'(1) \).

Key Concepts

Approximating Derivatives

Approximating derivatives is a fundamental concept in calculus that provides insights into a function's behavior around a particular point. When we talk about derivatives, we are referring to the rate at which a function is changing at any given point. Approximating a derivative involves calculating the average rate of change between two points close to one another on the curve of a function.

This is often achieved using difference quotients. A difference quotient is a formula used to estimate the slope of a secant line between two points on a graph. By varying the distance between these two points, we can approximate the instantaneous rate of change, or the derivative at a specific point.

The closer the points are to one another, the more accurate the approximation of the derivative becomes. This is because it more closely resembles the tangent line, which exactly represents the derivative at that point. Therefore, approximating derivatives using difference quotients is crucial in analyzing the nature of functions in calculus.

First Derivative Approximation

The first derivative approximation provides insight into the function's behavior at a specific point, usually represented as \( f'(x) \). In the original exercise, three different difference quotients are used to approximate \( f'(1) \):

• \( \frac{f(3)-f(1)}{3-1} = 0.04 \)
• \( \frac{f(2)-f(1)}{2-1} = 0.22 \)
• \( \frac{f(2)-f(0)}{2-0} = 0.88 \)

Each of these quotients represents an average rate of change over different intervals. The second difference quotient \( \frac{f(2) - f(1)}{2 - 1} \) directly involves \( x = 1 \), making its approximation potentially more reliable for \( f'(1) \).

The reason this approximation is likely more accurate is that it uses points closest to \( x = 1 \). This minimizes the errors that arise from larger intervals, where the secant line may deviate more significantly from the actual tangent line.

Accurate first derivative approximations are critical in understanding how functions behave over small changes, providing insights into variables' sensitivity and predicting future trends.

Calculus Problem Solving

Calculus problem solving often involves techniques to approximate and understand the functions via derivatives and integrals. In the specific context of the original exercise, students are tasked to analyze difference quotients to ascertain their proximity to the true derivative. This process involves:

• Identifying suitable intervals for approximation.
• Calculating difference quotients using function values.
• Comparing these approximations to determine their accuracy and relevance.

Engaging in these steps develops a deeper understanding of the function's local behavior around specific points.

Beyond just computing difference quotients, problem solving in calculus fosters analytical skills. Students learn to verify which approximations are more accurate by considering both the mathematical results and visual inspection from graphs.

In essence, calculus problem solving is about merging mathematical techniques with logical reasoning to uncover insights about functions, enabling us to tackle complex real-world problems effectively.