Problem 31 - 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 \(|\Delta f-d f|\) $$ f(x)=x^{2}+2 x, \quad x_{0}=1, \quad d x=0.1 $$

Step-by-Step Solution

Answer

The error is 0.01.

Step 1: Identify Components

The function given is \( f(x) = x^2 + 2x \). We are also given \( x_0 = 1 \) and \( dx = 0.1 \). We need to find \( \Delta f \), \( df \), and the error \( |\Delta f - df| \).

Step 2: Calculate \( \Delta f \)

To calculate the change in \( f \), we'll use \( \Delta f = f(x_0 + dx) - f(x_0) \). First, calculate \( f(x_0) = f(1) = 1^2 + 2 \times 1 = 3 \). Then, calculate \( f(x_0 + dx) = f(1.1) = 1.1^2 + 2 \times 1.1 = 1.21 + 2.2 = 3.41 \). Thus, \( \Delta f = 3.41 - 3 = 0.41 \).

Step 3: Calculate Derivative \( f'(x) \)

The derivative of \( f(x) = x^2 + 2x \) is \( f'(x) = 2x + 2 \).

Step 4: Calculate \( df \)

Use the derivative to calculate the estimate \( df = f'(x_0) \cdot dx \). Substitute \( x_0 = 1 \) into the derivative to get \( f'(1) = 2 \times 1 + 2 = 4 \). Then, \( df = 4 \times 0.1 = 0.4 \).

Step 5: Calculate Approximation Error

To find the approximation error, use \( |\Delta f - df| \). We have \( \Delta f = 0.41 \) and \( df = 0.4 \), so \( |0.41 - 0.4| = 0.01 \).

Key Concepts

Derivative

The derivative of a function is one of the fundamental concepts in differential calculus. It measures how a function changes as its input changes. In simple terms, the derivative represents the slope of the function at any point. For a given function \( f(x) = x^2 + 2x \), the derivative \( f'(x) \) can be found by applying standard differentiation rules.
In this case, the power rule and the constant multiple rule are used. This means you differentiate each term separately:

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( 2x \) is \( 2 \).

Combined, the derivative is \( f'(x) = 2x + 2 \). This indicates that at every point \( x \) on the curve, the slope is \( 2x + 2 \). Calculating the derivative is crucial because it provides insight into how the function behaves over small changes in \( x \). This knowledge is important for understanding concepts like function change and approximation.

Function Change

When understanding how a function changes, consider how the output \( f(x) \) moves as the input \( x \) shifts. In the exercise, you are asked to calculate \( \Delta f = f(x_0 + dx) - f(x_0) \). This means finding how much the function value has changed as \( x \) goes from \( x_0 \) to \( x_0 + dx \).

In our example, \( x_0 = 1 \) and \( dx = 0.1 \). First, you'll find \( f(1) \) by substituting 1 into our function. This results in \( f(1) = 3 \). Next, evaluate \( f(1.1) \) to get \( f(1.1) = 3.41 \). Thus, \( \Delta f \) is the difference between these two results:

• \( f(x_0 + dx) = 3.41 \)
• \( f(x_0) = 3 \)
• \( \Delta f = 3.41 - 3 = 0.41 \)

Understanding \( \Delta f \) helps measure exactly how the function's output shifts with a small input change.

Approximation Error

The approximation error is the difference between the actual change in the function \( \Delta f \) and the estimated change \( df \). This is calculated as \( |\Delta f - df| \). An approximation error gives us insight into how accurate our linear approximation is.

The goal is to see how close the linear estimation (based on the derivative) is to the actual change. In the example, after finding \( \Delta f = 0.41 \) and \( df = 0.4 \), the approximation error is \( |0.41 - 0.4| = 0.01 \).
This shows that the linear approximation is very accurate in this scenario, revealing that our derivative-based estimation closely predicts the real change in the function value. Understanding approximation error is fundamental when approximating functions, especially for predicting small changes.

Linear Approximation

Linear approximation is a method used to estimate the value of a function using its tangent line. It provides a simple way to approximate complex functions near a specific point. The function's derivative is the foundation for creating this linear approximation, as it defines the slope or rate of change at that point.
For a given function \( f(x) \), the linear approximation around \( x_0 \) can be described by:

• \( df = f'(x_0) \cdot dx \)

This formula predicts how the function value will change with a small change \( dx \).
In our case, after calculating \( f'(x) = 2x + 2 \) and substituting \( x_0 = 1 \), we find \( df = 0.4 \).
Linear approximation simplifies complex functions by reducing them to simple lines around small sections. This is extremely useful for calculations in mathematics and real-world applications where exact values are not necessary.