Problem 39 - 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) $$f(x)=x^{2}+2 x, \quad x_{0}=1, \quad d x=0.1$$

Step-by-Step Solution

Answer

\( \Delta f = 0.41 \), \( d f = 0.4 \), Error = 0.01

Step 1: Calculate \( \Delta f \)

To find the change in \( f(x) \), calculate \( f(x_0+dx) \) and \( f(x_0) \). \( f(x) = x^2 + 2x \). Given \( x_0 = 1 \) and \( dx = 0.1 \), calculate \[ f(x_0) = f(1) = 1^2 + 2 \times 1 = 3 \]\[ f(x_0 + dx) = f(1.1) = (1.1)^2 + 2 \times 1.1 = 1.21 + 2.2 = 3.41 \]Thus, \[ \Delta f = f(x_0 + dx) - f(x_0) = 3.41 - 3 = 0.41 \]

Step 2: Calculate the estimate \( d f \)

The estimated change \( d f \) is given by \[ d f = f'(x_0) \times dx \]. First, find the derivative of \( f(x) \): \( f(x) = x^2 + 2x \) so \( f'(x) = 2x + 2 \).Evaluate \( f'(x_0) \) at \( x_0 = 1 \): \[ f'(1) = 2 \times 1 + 2 = 4 \]Then calculate \[ d f = 4 \times 0.1 = 0.4 \]

Step 3: Calculate the approximation error \( |\Delta f - d f| \)

The error is the absolute value of the difference between the actual change and the estimated change: \[ |\Delta f - d f| = |0.41 - 0.4| = 0.01 \]

Key Concepts

Derivative Analysis

Derivative analysis is a fundamental concept in differential calculus that helps us understand how a function changes. It involves calculating the derivative, which represents the rate of change or the slope of the function at any point. For a function like \( f(x) = x^2 + 2x \), the derivative is calculated as \( f'(x) = 2x + 2 \). This expression tells us how steep the graph of the function is at each point.
Let's break it down:

• The term \( 2x \) indicates how the rate of change varies with \( x \). As \( x \) increases, the effect of this term grows.
• The constant term \( +2 \) indicates a fixed slope contributing to the rate of change all along the curve, no matter where you are on the \( x \)-axis.

Analyzing derivatives enables us to estimate how a small change in \( x \) (like \( dx = 0.1 \)) will affect the function's value, which is crucial in approximation methods.

Function Change

The change in a function, denoted as \( \Delta f \), represents how the function value alters when \( x \) progresses from some initial value \( x_0 \) to \( x_0 + dx \). It's essentially tracking the shift in the function's output.
In our example, with \( f(x) = x^2 + 2x \), we considered \( x_0 = 1 \) and \( dx = 0.1 \). Follow these steps:

• First, find the initial function value at \( x_0 = 1 \): \( f(1) = 3 \).
• Next, calculate the function at \( x_0 + dx = 1.1 \): \( f(1.1) = 3.41 \).
• The function change is \( \Delta f = 3.41 - 3 = 0.41 \).

This analysis is vital when we want to know the actual change between two points.

Approximation Error

The approximation error is an important concept in calculus that measures how much our estimate deviates from the true change. It's expressed as \( |\Delta f - df| \), which means the absolute difference between the actual and estimated change.
In our case:

• We calculated \( \Delta f = 0.41 \), the actual change in the function.
• The estimated change \( df \) using the derivative was \( 0.4 \).
• The approximation error is \( |0.41 - 0.4| = 0.01 \).

By understanding this error, we can assess the accuracy of our predictions when estimating how much a function will change with a small displacement in \( x \). This insight helps adjust models to reduce errors in future predictions.

Mathematical Estimation

Mathematical estimation using derivatives allows us to predict the outcome of a small change in a function's input efficiently. In calculus, we see this in the form of estimates like \( df = f'(x_0)\cdot dx \).
For our function:

• The derivative at \( x_0 = 1 \) is \( f'(1) = 4 \), indicating a steep rate of change.
• For a small change \( dx = 0.1 \), the estimated function change is \( 0.4 \).

This estimation approach is beneficial because it simplifies complex calculations by focusing on linear approximations, which are quicker to compute and often sufficiently accurate for small changes, reducing computational complexity and time.