Problem 33 - Thomas Calculus

Question

In Exercises \(31-36,\) 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 \(|\Delta f-d f|\) $$ f(x)=x^{3}-x, \quad x_{0}=1, \quad d x=0.1 $$

Step-by-Step Solution

Answer

∆f = 0.231, df = 0.2, error = 0.031.

Step 1: Calculate the Change ∆f

To find the change \( \Delta f \), we need to evaluate the function at two points and find their difference. First, calculate the value of the function at \( x_0 + dx = 1 + 0.1 = 1.1 \), which is:\[ f(1.1) = (1.1)^3 - 1.1 = 1.331 - 1.1 = 0.231 \]Then, compute the function at \( x_0 = 1 \):\[ f(1) = 1^3 - 1 = 0 \]The change \( \Delta f \) is:\[ \Delta f = 0.231 - 0 = 0.231 \]

Step 2: Determine the Estimate df

The estimate \( df \) involves finding the derivative of the function, then multiplying it by \( dx \). The derivative of \( f(x) = x^3 - x \) is:\[ f'(x) = 3x^2 - 1 \]Evaluate this derivative at \( x_0 = 1 \):\[ f'(1) = 3(1)^2 - 1 = 2 \]Now, multiply by \( dx = 0.1 \):\[ df = 2 \times 0.1 = 0.2 \]

Step 3: Calculate the Approximation Error

The approximation error is the absolute value of the difference between \( \Delta f \) and \( df \). From previous steps, \( \Delta f = 0.231 \) and \( df = 0.2 \). Thus, the error is:\[ |\Delta f - df| = |0.231 - 0.2| = 0.031 \]

Key Concepts

Function Evaluation

Function evaluation is the process of calculating the value of a function at a specific point. This involves substituting the value of the variable into the function and simplifying the expression to find the result.
In our exercise, we are tasked with evaluating the function \( f(x) = x^3 - x \) at two different points. First, consider the point \( x_0 = 1 \), which makes the evaluation simple:

• \( f(1) = 1^3 - 1 = 0 \)

Next, we evaluate the function at \( x = 1.1 \), which means substituting \( 1.1 \) into the function:

• \( f(1.1) = (1.1)^3 - 1.1 = 1.331 - 1.1 = 0.231 \)

These evaluations are the crucial steps needed to calculate the change in the function \( \Delta f \).
By understanding function evaluation, you gain insight into how the function behaves as its input changes, which is a fundamental concept in calculus.

Derivative Calculation

Calculating derivatives is essential for understanding how functions change. A derivative represents an instantaneous rate of change, providing critical information about a function's behavior at any given point.
In our specific problem, we consider the function \( f(x) = x^3 - x \). To find its derivative, apply the power rule for each term. The derivative \( f'(x) \) is given by:

• \( f'(x) = 3x^2 - 1 \)

We then evaluate this derivative at the specific point \( x_0 = 1 \):

• \( f'(1) = 3 \times 1^2 - 1 = 2 \)

The derivative tells us how quickly the function \( f \) is changing at \( x = 1 \), making it a critical component in estimating small changes through differential approximation.

Approximation Error

Approximation error quantifies the difference between the actual change in a function and the estimated change using differentials. It provides a measure of how accurate our differential approximation is.
In the exercise, we first found the actual change \( \Delta f \) to be 0.231, and the estimated change \( df \) to be 0.2. The approximation error is calculated as the absolute difference between these values:

• \( |\Delta f - df| = |0.231 - 0.2| = 0.031 \)

This error shows us that there is a small discrepancy between the estimated and actual changes. Understanding approximation error is essential for evaluating the reliability of derivative-based approximations, especially important in engineering and scientific calculations.

Differential Approximation

Differential approximation is a powerful method used to estimate small changes in functions, relying on derivatives to provide a linear approximation. It's based on the idea that over tiny intervals, functions behave almost like straight lines.
In the context of our problem, we use the derivative \( f'(x) = 3x^2 - 1 \) evaluated at \( x_0 = 1 \) to find that \( f'(1) = 2 \).

• By multiplying this value by the small change \( dx = 0.1 \), we get the differential approximation: \( df = 2 \times 0.1 = 0.2 \).

Differential approximation gives us a quick way to predict how the function \( f(x) \), changes as \( x \) shifts slightly from its initial value. While it might not always give the exact change, it provides a valuable approximation that is especially useful in cases where exact calculations are infeasible.