Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(F\) be any function whose domain contains \(-x\) whenever it contains \(x\). Prove each of the following. (a) \(F(x)-F(-x)\) is an odd function. (b) \(F(x)+F(-x)\) is an even function. (c) \(F\) can always be expressed as the sum of an odd and an even function.

Step-by-Step Solution

Answer

(a) Odd; (b) Even; (c) Sum of odd and even functions.

Step 1: Definition of an Odd Function

A function \( f(x) \) is called odd if \( f(-x) = -f(x) \) for all \( x \) in its domain. We need to verify this property for \( F(x) - F(-x) \).

Step 2: Show F(x) - F(-x) is Odd

Consider \( G(x) = F(x) - F(-x) \). Substitute \( -x \) into \( G \): \( G(-x) = F(-x) - F(--x) = F(-x) - F(x) = -(F(x) - F(-x)) = -G(x) \). Thus, \( G(x) \) satisfies the property of an odd function.

Step 3: Definition of an Even Function

A function \( f(x) \) is called even if \( f(-x) = f(x) \) for all \( x \) in its domain. We now need to verify this property for \( F(x) + F(-x) \).

Step 4: Show F(x) + F(-x) is Even

Consider \( H(x) = F(x) + F(-x) \). Substitute \( -x \) into \( H \): \( H(-x) = F(-x) + F(--x) = F(-x) + F(x) = H(x) \). Thus, \( H(x) \) satisfies the property of an even function.

Step 5: Express F as the Sum of Odd and Even Functions

From Steps 2 and 4, we have shown \( F(x) - F(-x) \) is odd and \( F(x) + F(-x) \) is even. Note that the function \( F(x) \) can be rewritten as \( F(x) = \frac{F(x) + F(-x)}{2} + \frac{F(x) - F(-x)}{2} \), where the first term is even and the second term is odd.

Key Concepts

Function Domain

The domain of a function is the collection of all possible inputs for which the function is defined. It is crucial to understand this concept when dealing with odd and even functions. For a function like \( F(x) \), which might be odd or even, the domain must include both \( x \) and \( -x \) for every \( x \) in the domain.
This ensures that when we substitute \( -x \) into the function, we aren't stepping outside of its valid input range.
In this exercise, understanding the domain is the first step to proving whether a function is odd or even.

Odd Function

An odd function is one that satisfies the property \( f(-x) = -f(x) \) for all values \( x \) in its domain.
This means that the graph of the function is symmetric about the origin.
In the exercise, we demonstrate this property by considering the function \( G(x) = F(x) - F(-x) \).
By substituting \( -x \), we find \( G(-x) = -G(x) \).
This shows that \( G(x) \) is indeed an odd function. This symmetry of functions is essential in various mathematical proofs and applications.

Even Function

An even function satisfies the condition \( f(-x) = f(x) \) for all \( x \) within the function's domain.
A key characteristic of even functions is their symmetry about the y-axis.
In our exercise, we need to confirm that \( H(x) = F(x) + F(-x) \) exhibits this property.
By substituting \( -x \), we obtain \( H(-x) = H(x) \), thus confirming \( H(x) \) is an even function.
Recognizing and utilizing functions' symmetries can simplify complex calculations significantly.

Function Decomposition

Function decomposition is the process of breaking down a complex function into simpler parts.
This is especially helpful in mathematical reasoning and problem-solving.
In the given exercise, we successfully decompose \( F(x) \) into the sum of an odd and even function.
Using the relationships previously established, \( F(x) \) can be expressed as:

• \( \frac{F(x) + F(-x)}{2} \)
• \( \frac{F(x) - F(-x)}{2} \)

The first term is even, and the second is odd.
This decomposition shows the beauty and utility of odd and even functions in mathematics.