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Chapter 12: Problem 2

Since \(f(-x)=-x \cos (-x)=-x \cos x=-f(x), f(x)\) is an odd function.

Short Answer

Expert verified
Yes, \( f(x) = x \cos x \) is an odd function because \( f(-x) = -f(x) \).

Step by step solution

01

Define the function

First, let's clearly define the function in question. According to the problem statement, the function is defined as \( f(x) = x \cos x \).
02

Substitute \( -x \) into the function

To determine if the function is odd, we substitute \( -x \) into the function: \[ f(-x) = -x \cos (-x). \]
03

Simplify \( \cos(-x) \)

Recall that cosine is an even function, meaning \( \cos(-x) = \cos(x) \). Therefore, the expression becomes: \[ f(-x) = -x \cos(x). \]
04

Compare \( f(-x) \) to \( -f(x) \)

The function \( f(x) \) can be expressed as \( f(x) = x \cos x \). Its negative \( -f(x) \) is: \[ -f(x) = -(x \cos x) = -x \cos x. \]Compare this with the expression for \( f(-x) \) obtained in Step 3: \[ f(-x) = -x \cos x. \] Since \( f(-x) = -f(x) \), this verifies that \( f(x) \) is indeed an odd function.

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Key Concepts

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

Trigonometric Identity
Trigonometric identities are essential tools in mathematics, helping us manipulate and simplify expressions involving trigonometric functions. One key trigonometric identity is the property of the cosine function:
  • Cosine is an even function. This means for any angle \( x \), the identity \( \cos(-x) = \cos(x) \) holds true.
This property is crucial in understanding the behavior of functions involving cosines, especially when determining if a function is odd or even.

Remember, an even function, like cosine, retains its value when the input is negated. This identity simplifies many calculations and is a fundamental aspect of trigonometry required in various math problems, from basic equations to complex integrals.
Even and Odd Functions
In mathematics, functions can often be classified as either even or odd based on their symmetry properties.
  • An even function satisfies the condition: \( f(-x) = f(x) \) for all values of \( x \) in its domain.
  • An odd function satisfies the condition: \( f(-x) = -f(x) \).
The symmetry properties of these functions can help simplify complex mathematical problems and analyses. Odd functions have specific symmetry around the origin, often reflected graphically as rotational symmetry.

The function \( f(x) = x \cos(x) \) is determined to be an odd function. This conclusion arises from substituting \(-x\) into the function and showing that the result is the negative of \( f(x) \). Understanding these properties assists in solving equations and transforming graphs efficiently.
Function Transformation
Function transformations involve altering the appearance or form of a function graphically without changing its basic shape. These transformations include operations like shifts, stretches, compressions, and reflections.
  • Shift: Moving the graph horizontally or vertically.
  • Stretch/Compression: Modifying the graph's steepness or width.
  • Reflection: Flipping the graph across an axis.
With trigonometric functions like \( f(x) = x \cos(x) \), transformations often help in understanding its periodicity and oscillatory behavior.

Identifying the transformations involved can aid in predicting the graph's reaction to different input values. This understanding is not only vital in graph theory but is also applicable in fields such as engineering and physics, where modeling wave-like phenomena is required.

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Most popular questions from this chapter

The eigenfunctions are \(\sin \alpha_{n} x\) where \(\tan \alpha_{n}=-\alpha_{n} .\) Thus $$\begin{aligned} \left\|\sin \alpha_{n} x\right\|^{2} &=\int_{0}^{1} \sin ^{2} \alpha_{n} x d x=\frac{1}{2} \int_{0}^{1}\left(1-\cos 2 \alpha_{n} x\right) d x \\ &=\left.\frac{1}{2}\left(x-\frac{1}{2 \alpha_{n}} \sin 2 \alpha_{n} x\right)\right|_{0} ^{1}=\frac{1}{2}\left(1-\frac{1}{2 \alpha_{n}} \sin 2 \alpha_{n}\right) \\ &=\frac{1}{2}\left[1-\frac{1}{2 \alpha_{n}}\left(2 \sin \alpha_{n} \cos \alpha_{n}\right)\right] \\ &=\frac{1}{2}\left[1-\frac{1}{\alpha_{n}} \tan \alpha_{n} \cos \alpha_{n} \cos \alpha_{n}\right] \\ &=\frac{1}{2}\left[1-\frac{1}{\alpha_{n}}\left(-\alpha_{n} \cos ^{2} \alpha_{n}\right)\right]=\frac{1}{2}\left(1+\cos ^{2} \alpha_{n}\right). \end{aligned}$$

Since \(f(x)\) is not defined for \(x<0\), it is neither even nor odd.

$$\begin{array}{l}a_{0}=\frac{2}{\pi} \int_{0}^{\pi} \sin x d x=\frac{4}{\pi} \\\a_{n}=\frac{2}{\pi} \int_{0}^{\pi} \sin x \cos n x d x=\frac{1}{\pi} \int_{0}^{\pi}[\sin (n+1) x-\sin (n-1) x] d x=\frac{2\left[(-1)^{n}+1\right]}{\pi\left(1-n^{2}\right)} \quad \text { for } n=2,3,4, \ldots \\\b_{n}=\frac{2}{\pi} \int_{0}^{\pi} \sin x \sin n x d x=\frac{1}{\pi} \int_{0}^{\pi}[\cos (n-1) x-\cos (n+1) x] d x=0 \quad \text { for } n=2,3,4, \ldots\end{array}$$ $$\begin{array}{l}a_{1}=\frac{1}{\pi} \int_{0}^{\pi} \sin 2 x d x=0 \\\b_{1}=\frac{2}{\pi} \int_{0}^{\pi} \sin ^{2} x d x=1 \\\f(x)=\sin x \\\f(x)=\frac{2}{\pi}+\frac{2}{\pi} \sum_{n=2}^{\infty} \frac{(-1)^{n}+1}{1-n^{2}} \cos n x \end{array}$$

Since \(f(-x)=e^{-x}-e^{x}=-f(x), f(x)\) is an odd function.

$$\begin{array}{l}a_{0}=2 \int_{0}^{1}\left(x^{2}+x\right) d x=\frac{5}{3} \\\a_{n}=2 \int_{0}^{1}\left(x^{2}+x\right) \cos n \pi x d x=\left.\frac{2\left(x^{2}+x\right)}{n \pi} \sin n \pi x\right|_{0} ^{1}-\frac{2}{n \pi} \int_{0}^{1}(2 x+1) \sin n \pi x d x=\frac{2}{n^{2} \pi^{2}}\left[3(-1)^{n}-1\right] \\\b_{n}=2 \int_{0}^{1}\left(x^{2}+x\right) \sin n \pi x d x=-\left.\frac{2\left(x^{2}+x\right)}{n \pi} \cos n \pi x\right|_{0} ^{1}+\frac{2}{n \pi} \int_{0}^{1}(2 x+1) \cos n \pi x d x\end{array}$$ $$\begin{array}{l}=\frac{4}{n \pi}(-1)^{n+1}+\frac{4}{n^{3} \pi^{3}}\left[(-1)^{n}-1\right] \\\f(x)=\frac{5}{6}+\sum_{n=1}^{\infty} \frac{2}{n^{2} \pi^{2}}\left[3(-1)^{n}-1\right] \cos n \pi x \\\f(x)=\sum_{n=1}^{\infty}\left(\frac{4}{n \pi}(-1)^{n+1}+\frac{4}{n^{3} \pi^{3}}\left[(-1)^{n}-1\right]\right) \sin n \pi x\end{array}$$
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