Problem 42 - 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) $$f(x)=x^{4}, \quad x_{0}=1, \quad d x=0.1$$

Step-by-Step Solution

Answer

a. \( \Delta f = 0.4641 \); b. \( df = 0.4 \); c. Error = \( 0.0641 \).

Step 1: Calculate Change in Function Value

The change in the function value, denoted as \( \Delta f \), is given by \( f(x_0 + dx) - f(x_0) \). Substitute \( x_0 = 1 \) and \( dx = 0.1 \) into the function \( f(x) = x^4 \). 1. Calculate \( f(1) = 1^4 = 1 \).2. Calculate \( f(1 + 0.1) = (1.1)^4 \).3. \( (1.1)^4 = 1.1 \times 1.1 \times 1.1 \times 1.1 = 1.4641 \).4. Therefore, \( \Delta f = 1.4641 - 1 = 0.4641 \).

Step 2: Calculate Differential Change

The differential change, denoted as \( df \), is given by \( f'(x_0) dx \). First, we need the derivative of \( f(x) = x^4 \).1. The derivative is \( f'(x) = 4x^3 \).2. Substitute \( x_0 = 1 \) into \( f'(x) \): \( f'(1) = 4 \times 1^3 = 4 \).3. Therefore, \( df = 4 \times 0.1 = 0.4 \).

Step 3: Calculate Approximation Error

The approximation error is the absolute value of the difference between the actual change \( \Delta f \) and the differential change \( df \). It is calculated as \( |\Delta f - df| \).1. Substitute \( \Delta f = 0.4641 \) and \( df = 0.4 \).2. Calculate the absolute difference: \(| \Delta f - df | = |0.4641 - 0.4| = 0.0641 \).

Key Concepts

Function Change

Whenever a function is subjected to a change in its input, there is a corresponding change in its output. This is known as "Function Change". For the given function in our problem, we considered how the value changes when the input changes from an initial point, denoted as \(x_0\), to \(x_0 + dx\).
For example, if we have \(f(x) = x^4\), changing \(x\) from \(1\) to \(1.1\) results in a function change calculated as:

• Find \(f(x_0) = f(1) = 1^4 = 1\).
• Calculate \(f(x_0 + dx) = f(1.1) = (1.1)^4 = 1.4641\).
• Thus, the change in function value \(\Delta f = f(1.1) - f(1) = 1.4641 - 1 = 0.4641\).

Understanding this step helps lay the foundation for deeper concepts in calculus, showing how small changes in input affect the function.

Differential Approximation

Differential Approximation is a powerful tool in calculus, enabling us to estimate how a function behaves at a small neighborhood around a specific point.
To approximate, we rely on the derivative of the function, which gives us the slope or rate of change at a point \(x_0\). For our function \(f(x) = x^4\), the derivative is:

• Compute the derivative: \(f'(x) = 4x^3\).
• Substitute \(x_0 = 1\) into the derivative: \(f'(1) = 4 \times 1^3 = 4\).
• Estimate the differential change: \(df = f'(1) \cdot dx = 4 \times 0.1 = 0.4\).

This approximated value \(df\) attempts to predict what \(\Delta f\) should be given a small change \(dx\). In many real-world situations, this approximation simplifies complex calculations.

Approximation Error

Despite its usefulness, differential approximation is not always perfect. The difference between the precise change in the function, \(\Delta f\), and its estimated change, \(df\), is called the "Approximation Error".
This error quantifies the accuracy of our estimation and is essential for understanding how well our approximation fits:

• Calculate the actual change \(\Delta f = 0.4641\).
• Determine the differential approximation value \(df = 0.4\).
• Find the approximation error: \(|\Delta f - df| = |0.4641 - 0.4| = 0.0641\).

The smaller this error, the closer our approximation is to reality. This is vital in applications where precision is crucial, like engineering and physics, helping to ensure models and predictions are reliable.