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Chapter 3: Problem 31

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 \(|\Delta f-d f|\). $$f(x)=x^{3}-x, \quad x_{0}=1, \quad d x=0.1$$

Short Answer

Expert verified
The change is 0.231, the estimate is 0.2, and the error is 0.031.

Step by step solution

01

Calculate the Change in the Function Value

We are given the function \( f(x) = x^3 - x \). We need to find \( \Delta f = f(x_0 + dx) - f(x_0) \). First, substitute \( x_0 = 1 \) and \( dx = 0.1 \): \( f(x_0) = f(1) = 1^3 - 1 = 0 \). Next, calculate \( f(x_0 + dx) = f(1 + 0.1) = f(1.1) = (1.1)^3 - 1.1 \). Calculate \( (1.1)^3 = 1.331 \), so \( f(1.1) = 1.331 - 1.1 = 0.231 \). Thus, \( \Delta f = f(1.1) - f(1) = 0.231 - 0 = 0.231 \).
02

Calculate the Estimate Using the Derivative

Find the derivative of \( f(x) = x^3 - x \), which is \( f'(x) = 3x^2 - 1 \). Evaluate the derivative at \( x_0 = 1 \): \( f'(1) = 3(1)^2 - 1 = 2 \). Use this derivative to estimate \( df = f'(x_0)\,dx = 2 \times 0.1 = 0.2 \).
03

Calculate the Approximation Error

The approximation error is given by \( |\Delta f - df| \). From Step 1, \( \Delta f = 0.231 \) and from Step 2, \( df = 0.2 \). Thus, the error is \( |0.231 - 0.2| = 0.031 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Calculus
Differential Calculus plays a key role in understanding how functions change. It is mainly concerned with the concept of the derivative, which provides us with a good estimate of how a function value changes as its input shifts slightly. In this context, the derivative acts like a powerful tool to predict how functions behave near a particular point.
  • When we have a function like \( f(x) = x^3 - x \), calculating its derivative helps us understand the function's rate of change.
  • The derivative, denoted \( f'(x) \), is found using rules like the power rule, which, for a power of \( n \), states \( (x^n)' = n\cdot x^{n-1} \).
  • For our example, this calculation leads to \( f'(x) = 3x^2 - 1 \), providing insights about how the function behaves for small changes in \( x \).
The concept becomes practical when you wish to approximate the change in a function when \( x \) nudges from \( x_0 \) to \( x_0 + dx \). This derivative-driven estimate is called the "differential" \( df \), representing an easy way to approximate that change efficiently.
Function Analysis
Function Analysis involves examining a function's properties and behavior. It involves assessments like finding the function's values at given points and understanding the function's continuity and differentiability.
  • For the function \( f(x) = x^3 - x \), the value \( f(x_0) \) at a point \( x_0 = 1 \) is given by substituting the point: \( f(1) = 1^3 - 1 = 0 \).
  • Similarly, to analyze the change as \( x \) moves slightly, compute \( f(x_0 + dx) = f(1 + 0.1) = (1.1)^3 - 1.1 = 0.231 \).
Here, understanding how the function's value changes, \( \Delta f = f(x_0 + dx) - f(x_0) \), allows us to quantify this change precisely. It is a simple subtraction providing the true change in the function value over a small interval. This part of function analysis ensures that we accurately capture how the function behaves near any given point.
Error Approximation
Error Approximation is crucial for understanding how close our derivative-based estimates are to the actual change in a function. While the differential \( df \) offers an estimate for \( \Delta f \), it may not be exact. The error approximation helps highlight the difference between these two values.
  • In our problem, we calculated \( \Delta f = 0.231 \) and \( df = 0.2 \), showcasing the importance of evaluating the discrepancy \(|\Delta f - df| = 0.031 \).
  • This error value indicates how much precision we lose by relying solely on the derivative-based estimate, helping us understand the limitations of using \( df \).
Error approximation is particularly useful in practical applications where knowing the potential error margin is critical. It informs us about the accuracy and reliability of our estimates, ensuring we have a realistic view of how the approximation aligns with actual changes.

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Most popular questions from this chapter

Is there a value of \(b\) that will make $$g(x)=\left\\{\begin{array}{ll} x+b, & x < 0 \\ \cos x, & x \geq 0 \end{array}\right.$$ continuous at \(x=0 ?\) Differentiable at \(x=0 ?\) Give reasons for your answers.

Find \(d y / d t\) $$y=\sin ^{2}(\pi t-2)$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1$$

In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in W眉rzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about \(7 \mathrm{L} / \mathrm{min} .\) At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min}\). If you are a trained marathon runner running a marathon, your cardiac output can be as high as \(30 \mathrm{L} / \mathrm{min}\). Your cardiac output can be calculated with the formula $$y=\frac{Q}{D},$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\), \(y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}\), fairly close to the \(6 \mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u+\frac{1}{\cos ^{2} u}, \quad u=g(x)=\pi x, \quad x=1 / 4$$
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