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Chapter 4: Problem 63

Show that \(|\cos x-1| \leq|x|\) for all \(x\) -values.

Short Answer

Expert verified
The inequality \(|\cos x-1| \leq|x|\) holds for all \(x\).

Step by step solution

01

Understand the Function Behavior

We start by understanding the behavior of the function \(|\cos x - 1|\). The cosine function, \( \cos x \), is always between \(-1\) and \(1\) for all real \(x\). So \( \cos x - 1 \) ranges from \(-2\) to \(0\). Thus, \(|\cos x - 1|\) ranges from \(0\) to \(2\).
02

Use the Taylor Series Approximation

Consider the Taylor series expansion of \( \cos x \) around \(0\). The Taylor series is \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \). Thus, \( \cos x - 1 \approx -\frac{x^2}{2} \) for small values of \(x\). This shows that \(|\cos x - 1| \approx \frac{x^2}{2} \leq |x|\) for small \(x\).
03

Evaluate at Special Points

Check some specific values, such as \(x = 0\), where \( \cos(0) = 1 \) and hence \(|\cos(0)-1| = 0 \leq 0 = |0|\). For \(x = \pi\), \(\cos(\pi) = -1\), so \(|\cos(\pi) - 1| = 2\), but \(|\pi| > 2\). Thus inequality holds.
04

Verify the Inequality using Derivative

The function \( f(x) = \cos x - 1 + x \) and \( g(x) = \cos x - 1 - x \) should both be non-negative for the inequality to hold. Compute the derivatives: \( f'(x) = -\sin x + 1 \) and \( g'(x) = -\sin x - 1 \). Analyze where these derivatives are zero or non-positive to confirm these functions do not change sign improperly, showing \(f(x)\) does not fall below zero and \(g(x)\) does not rise above zero.

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Key Concepts

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

Trigonometric Functions
Trigonometric functions, including cosine (\( \cos x \)), play a critical role in analyzing periodic phenomena. The cosine function oscillates between \(-1\) and \(1\) for any real number \(x\). This range is what makes the cosine function predictable and highly useful in various applications like engineering and physics.
  • When dealing with the expression \(\cos x - 1\), we see that it shifts the cosine graph down by 1 unit, resulting in a range from \(-2\) to \(0\).
  • Thus, the absolute value expression \(|\cos x - 1|\) takes on values from 0 to 2, reflecting how far the result is from zero.

Understanding the motion and oscillation of trigonometric functions helps provide intuition into their behavior in inequalities and other mathematical problems.
Taylor Series
The Taylor series is a fundamental concept that assists in approximating complex functions using polynomials. It's like expressing the function as an infinite sum of terms made from its derivatives at a specific point.
  • In the case of \(\cos x\) around zero, the Taylor series is \(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\)
  • When considering \(\cos x - 1\), we approximate it using the first few terms: \(-\frac{x^2}{2}\) for small values of \(x\)

This approximation \(|\cos x - 1| \approx \frac{x^2}{2} \leq |x|\) is useful for small \(x\), providing a means to compare magnitudes easily.
  • The series provides the insight that for sufficiently small \(x\), \(|\cos x - 1|\) is dominated by \(\frac{x^2}{2}\)

This makes it easier to establish inequalities involving trigonometric functions near specific points.
Derivative Analysis
Derivatives are powerful tools in calculus that help understand how functions change, indicating where they increase, decrease, or remain constant.
  • In the exercise, we analyze the functions \( f(x) = \cos x - 1 + x \) and \( g(x) = \cos x - 1 - x \) to verify the inequality \(|\cos x - 1| \leq |x|\).
  • The derivative \( f'(x) = -\sin x + 1 \) helps determine where \( f(x) \) increases or decreases.
  • Similarly, \( g'(x) = -\sin x - 1 \) lets us analyze how \( g(x) \) behaves across its domain.

By exploring the sign and critical points of these derivatives, we can confirm that \( f(x) \) does not decrease past zero and \( g(x) \) does not increase beyond zero, ensuring that the inequality holds for all \(x\).
  • In essence, derivative analysis provides a way to verify function behavior without having to rely entirely on function evaluations.

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Most popular questions from this chapter

One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is $$A(q)=\frac{k m}{q}+c m+\frac{h q}{2}$$ where \(q\) is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), \(k\) is the cost of placing an order (the same, no matter how often you order), \(c\) is the cost of one item (a constant), \(m\) is the number of items sold each week (a constant), and \(h\) is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to find the quantity that will minimize \(A(q) .\) What is it? (The formula you get for the answer is called the Wilson lot size formula.) b. Shipping costs sometimes depend on order size. When they do, it is more realistic to replace \(k\) by \(k+b q,\) the sum of \(k\) and a constant multiple of \(q\). What is the most economical quantity to order now?

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(\frac{1}{x^{2}}-x^{2}-\frac{1}{3}\right) d x$$

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(-3 \csc ^{2} x\right) d x$$

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(8 y-\frac{2}{y^{1 / 4}}\right) d y$$

Use a CAS to solve the initial value problems. Plot the solution curves.$$y^{\prime}=\frac{1}{x}+x, \quad y(1)=-1$$
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