Problem 36 - 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 \(|\Delta f-d f|\) $$ f(x)=x^{3}-2 x+3, \quad x_{0}=2, \quad d x=0.1 $$

Step-by-Step Solution

Answer

a. \( \Delta f = 1.061 \); b. \( df = 1.0 \); c. Error = \( 0.061 \).

Step 1: Identify the Function and Values

We are given the function \( f(x) = x^3 - 2x + 3 \), initial value \( x_0 = 2 \), and increment \( dx = 0.1 \). Our task involves three calculations: change in \( f \), differential of \( f \), and error estimation.

Step 2: Calculate the Change in f (Δf)

The change \( \Delta f \) is given by \( f(x_0 + dx) - f(x_0) \). First, calculate \( f(x_0) = f(2) = 2^3 - 2\cdot2 + 3 = 8 - 4 + 3 = 7 \). Then, calculate \( f(x_0 + dx) = f(2.1) = (2.1)^3 - 2\cdot2.1 + 3 \). Using the binomial expansion for power 3: \( (2.1)^3 = 2.1 \cdot 2.1 \cdot 2.1 = 9.261 \). Then, \( f(2.1) = 9.261 - 4.2 + 3 = 8.061 \). Therefore, \( \Delta f = 8.061 - 7 = 1.061 \).

Step 3: Compute the Differential of f (df)

The differential \( df \) is estimated by \( f'(x_0) \cdot dx \). First, determine \( f'(x) = 3x^2 - 2 \). Thus, \( f'(2) = 3\cdot2^2 - 2 = 12 - 2 = 10 \). Now, \( df = 10 \cdot 0.1 = 1.0 \).

Step 4: Determine the Approximation Error |Δf - df|

The error is calculated as \( |\Delta f - df| = |1.061 - 1.0| = 0.061 \). This measures how close our linear approximation is to the actual change in \( f \).

Key Concepts

Function Change

In calculus, understanding how a function changes when its input changes is fundamental. For any function \( f(x) \), the change \( \Delta f \) is determined by how \( f \) reacts to a small modification in \( x \) from an initial value \( x_0 \) to a nearby value \( x_0 + dx \). This is expressed as:

• \( \Delta f = f(x_0 + dx) - f(x_0) \)

By substituting specific numbers into the function, you can determine its exact change. For instance, in our exercise, with \( f(x) = x^3 - 2x + 3 \), initial value \( x_0 = 2 \), and \( dx = 0.1 \), you first calculate \( f(x_0) \) and then \( f(x_0 + dx) \). The difference gives you \( \Delta f \). Why is this useful? Knowing \( \Delta f \) shows you the accurate change in the outcome of the function when \( x \) is slightly altered. This exact value is pivotal for precise assessments in various scenarios, like engineering and physics, where small variations can lead to significant impacts.

Differential Approximation

Differential Approximation simplifies the process of estimating how a function behaves when \( x \) varies slightly. Instead of finding the exact \( \Delta f \), which could be complex, it uses the function's derivative to provide an estimate, denoted as \( df \). This is calculated by:

• \( df = f'(x_0) \cdot dx \)

Here, \( f'(x) \) represents the derivative of the function, providing the rate of change of the function with respect to \( x \). For our specific function \( f(x) = x^3 - 2x + 3 \), the derivative \( f'(x) = 3x^2 - 2 \). By evaluating \( f'(x_0) \) at \( x_0 = 2 \) and multiplying by \( dx = 0.1 \), you get \( df \). This approximation comes in handy when you need quick estimations without diving into detailed calculations. It's extensively used in fields like economics and biology for prompt predictions where exact numbers aren't always necessary.

Error Estimation

Error Estimation focuses on the difference between the exact change \( \Delta f \) and the estimated change \( df \) from the differential approximation. The formula for error estimation is:

• \(|\Delta f - df| \)

In practice, this value tells you how far off the approximation \( df \) is from the real change \( \Delta f \). With our function example, it demonstrates the closeness of our approximation, computed as: \( |1.061 - 1.0| = 0.061 \). Understanding this error is crucial, especially in sensitive calculations where precision matters, such as computer science algorithms or financial models. It allows analysts and researchers to gauge the reliability of their approximations, ensuring decisions are made based on sound assessments.