Problem 44 - 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) $$f(x)=x^{3}-2 x+3, x_{0}=2, \quad d x=0.1$$

Step-by-Step Solution

Answer

a. Δf = 1.061; b. df = 1.0; c. Error = 0.061.

Step 1: Calculate f(x_0)

First, calculate the value of the function \( f(x) \) at \( x_0 = 2 \). With \( f(x) = x^3 - 2x + 3 \), we plug in \( x_0 \): \[ f(x_0) = 2^3 - 2(2) + 3 = 8 - 4 + 3 = 7 \] So, \( f(x_0) = 7 \).

Step 2: Calculate f(x_0 + dx)

Next, calculate \( f(x) \) at \( x_0 + dx = 2.1 \). Plug \( x = 2.1 \) into the function: \[ f(x_0 + dx) = (2.1)^3 - 2(2.1) + 3 \] Calculate: \[ (2.1)^3 = 9.261, \ -2(2.1) = -4.2 \] Therefore: \[ f(x_0 + dx) = 9.261 - 4.2 + 3 = 8.061 \] So, \( f(x_0 + dx) = 8.061 \).

Step 3: Find the Change Δf

The change in function value \( \Delta f \) is computed by subtracting the value at \( x_0 \) from the value at \( x_0 + dx \): \[ \Delta f = f(x_0 + dx) - f(x_0) = 8.061 - 7 = 1.061 \] Thus, \( \Delta f = 1.061 \).

Step 4: Calculate the Derivative f'(x)

Find the derivative \( f'(x) \) of the function \( f(x) = x^3 - 2x + 3 \): \[ f'(x) = 3x^2 - 2 \] Find \( f'(x) \) at \( x_0 = 2 \): \[ f'(2) = 3(2)^2 - 2 = 12 - 2 = 10 \] So, \( f'(x_0) = 10 \).

Step 5: Calculate the Estimate df

The estimate \( df \) is given by the product of the derivative \( f'(x_0) \) and \( dx \): \[ df = f'(x_0) \cdot dx = 10 \cdot 0.1 = 1.0 \] Thus, \( df = 1.0 \).

Step 6: Find the Approximation Error |Δf - df|

Finally, compute the approximation error, which is the absolute difference between \( \Delta f \) and \( df \): \[ |\Delta f - df| = |1.061 - 1.0| = 0.061 \] So, the approximation error is \( 0.061 \).

Key Concepts

Function Analysis

Function analysis in calculus involves examining how a function behaves, changes, and is structured. Here, we are dealing with a function given by

• \( f(x) = x^3 - 2x + 3 \)

To understand function changes, we investigate values of the function at specific points. In this exercise, the task was to see how the function value changes when \( x \) moves from \( x_0 = 2 \) to \( x_0 + dx = 2.1 \).To start, we determine the initial value of the function at \( x_0 = 2 \) which gives us \( f(x_0) = 7 \). Function analysis often involves calculating such values to see how a function changes or to understand its patterns, like its growth or decay at certain intervals.

Derivatives

Derivatives are a fundamental part of calculus that measures how a function changes at any given point. It represents the slope or rate of change of the function. In our exercise,

• we use the function \( f(x) = x^3 - 2x + 3 \).
• The derivative, denoted as \( f'(x) \), is given by \( f'(x) = 3x^2 - 2 \).

The derivative tells us how steeply the function values increase or decrease as \( x \) changes, giving an idea of the function's behavior.

To find the slope specifically at \( x_0 = 2 \), we calculated \( f'(2) = 10 \). This shows that at \( x = 2 \), the function's rate of change is 10, indicating a rapid increase.
Derivatives enable us to predict how small changes in \( x \) can affect \( f(x) \), a crucial aspect for solving real-world problems involving rates and optimizations.

Approximation Errors

Approximation errors occur when the calculated estimate slightly differs from the actual change. This can happen when measuring how a function changes over a tiny interval. In our task, the change in the function's value is illustrated by

• \( \Delta f = f(x_0 + dx) - f(x_0) = 1.061 \)
• The estimate of this change using derivatives, \( df = f'(x_0) \cdot dx = 1.0 \)

The approximation error is found as

• \( |\Delta f - df| = |1.061 - 1.0| = 0.061 \).

This shows how close our derivative-based prediction is to the actual functional change.

By understanding approximation errors, one can evaluate the accuracy of approximations used in calculations and predictions. This insight is vital in engineering, physics, and other fields where precise measurements influence outcomes.