Problem 41 - University Calculus: Early Transcendentals

Question

Each function \(f(x)\) changes value when \(x\) changes from \(x_{0}\) to \(x_{0}+d x .\) Find a. the change \(\Delta f=f\left(x_{0}+d x\right)-f\left(x_{0}\right)\) b. the value of the estimate \(d f=f^{\prime}\left(x_{0}\right) d x ;\) and c. the approximation error \(|\Delta f-d f|\) (GRAPH CANNOT COPY) $$f(x)=x^{3}-x, \quad x_{0}=1, \quad d x=0.1$$

Step-by-Step Solution

Answer

a. \(\Delta f = 0.231\), b. \(df = 0.2\), c. error = 0.031

Step 1: Calculate \( \Delta f \)

First, evaluate the function at \( x_0 + dx = 1.1 \): \[ f(1.1) = (1.1)^3 - 1.1 = 1.331 - 1.1 = 0.231 \] Then, evaluate the function at \( x_0 = 1 \): \[ f(1) = 1^3 - 1 = 0 \] Now, compute \( \Delta f \): \[ \Delta f = f(1.1) - f(1) = 0.231 - 0 = 0.231 \]

Step 2: Determine the derivative at \( x_0 \)

Find the derivative of the function: \[ f'(x) = 3x^2 - 1 \] Evaluate the derivative at \( x_0 = 1 \): \[ f'(1) = 3(1)^2 - 1 = 3 - 1 = 2 \]

Step 3: Compute \( df \)

Use the formula for the estimate \( df \): \[ df = f'(x_0) \cdot dx = 2 \cdot 0.1 = 0.2 \]

Step 4: Calculate the approximation error

Determine the absolute difference between \( \Delta f \) and \( df \):\[ |\Delta f - df| = |0.231 - 0.2| = 0.031 \]

Key Concepts

Differential Calculus

Differential Calculus is a branch of calculus focused on the study of how things change. It deals with the concept of derivatives, which describe the rate at which a function's value is changing at any given point.

• The main goal of differential calculus is to find the rate of change of a function with respect to one of its variables. This rate of change is called the derivative.
• The derivative provides a powerful tool to find instantaneous rates of change and slopes of tangent lines.

Whether you're calculating the velocity of a car or determining the decline in sales over time, differential calculus aids in understanding these continuous changes. In our exercise, we specifically delve into finding derivatives to approximate function changes.

Function Change

The change in a function's value, often denoted by \(\Delta f\), is a central idea in calculus. It represents how the function's output changes as the input changes. Essentially, it captures the difference between the function's values at two distinct points.

• In our exercise, the difference \(\Delta f = f(x_{0} + dx) - f(x_{0})\) shows how much the function value changes when \(x\) changes from \(x_{0}\) to \(x_{0} + dx\).
• This concept is useful for understanding real-world phenomena where input variables cause changes in outcomes.

These changes give important initial insights into the behavior of the function over small intervals.

Derivative

The derivative is a cornerstone in calculus. It tells us how a function's value changes as its input changes. Mathematically, it is defined as the limit of the ratio of the change in the function to the change in its input, as that change in input approaches zero.

• In our problem, we find the derivative of the function, \(f(x) = x^{3} - x\), which is \(f'(x) = 3x^{2} - 1\).
• By plugging in \(x_{0}=1\), we find the rate of change at that particular point, \(f'(1) = 2\).

This helps estimate small changes in the function's value when the input changes by a small amount.

Approximation Error

The approximation error gives us an idea of how accurate our derivative-based estimates are. This error is the difference between the actual change in the function's value, \(\Delta f\), and the estimated change, \(d f\).

• For our function, the approximation error is calculated as \(|\Delta f - d f| = |0.231 - 0.2| = 0.031\).
• This value of \(0.031\) shows how close our linear approximation (using the derivative) is to the actual change.

The approximation error provides a measure of precision in mathematical modeling, highlighting the importance of accurate estimates in predictions and analyses.