Question

The functions are neither even nor odd. Write each of them as the sum of an even function and an odd function. (a) \(x^{5}-x^{4}+x^{3}-1\) (b) \(1+e^{x}\)

Short answer

The function \f(x) = x^5 - x^4 + x^3 - 1\ can be written as \(-x^4 - 1 + x^5 + x^3\). The function \g(x) = 1 + e^x\ can be written as \1 + \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2}\.

Step-by-step solution

- Understand the Problem

Given functions are neither even nor odd. We need to write each function as the sum of an even function and an odd function.

- Formula for Even and Odd Components

A function can be decomposed into an even part and an odd part using the formulas:- Even part: \(\frac{f(x) + f(-x)}{2}\)- Odd part: \(\frac{f(x) - f(-x)}{2}\)

- Apply to (a) \(x^5 - x^4 + x^3 - 1\)

Let \(f(x) = x^5 - x^4 + x^3 - 1\).Calculate \(f(-x)\): \(f(-x) = (-x)^5 - (-x)^4 + (-x)^3 - 1 = -x^5 - x^4 - x^3 - 1\)Even part: \(\frac{f(x) + f(-x)}{2} = \frac{x^5 - x^4 + x^3 - 1 - x^5 - x^4 - x^3 - 1}{2} = -\frac{2x^4 + 2}{2} = -x^4 - 1\)Odd part: \(\frac{f(x) - f(-x)}{2} = \frac{x^5 - x^4 + x^3 - 1 + x^5 + x^4 + x^3 + 1}{2} = \frac{2x^5 + 2x^3}{2} = x^5 + x^3\)Thus, \f(x)\ can be written as \(f(x) = (-x^4 - 1) + (x^5 + x^3)\).

- Apply to (b) \(1 + e^x\)

Let \(g(x) = 1 + e^x\).Calculate \(g(-x)\): \(g(-x) = 1 + e^{-x}\)Even part: \(\frac{g(x) + g(-x)}{2} = \frac{1 + e^x + 1 + e^{-x}}{2} = \frac{2 + e^x + e^{-x}}{2} = 1 + \frac{e^x + e^{-x}}{2}\)Odd part: \(\frac{g(x) - g(-x)}{2} = \frac{1 + e^x - (1 + e^{-x})}{2} = \frac{e^x - e^{-x}}{2}\)Thus, \g(x)\ can be written as \(g(x) = \left(1 + \frac{e^x + e^{-x}}{2}\right) + \left(\frac{e^x - e^{-x}}{2}\right)\).

Key concepts

even function

An **even function** has a specific symmetry: it remains unchanged when you replace \(x\) with \(-x\). This means the graph of an even function is symmetric about the y-axis. For a function \(f(x)\) to be even, it must satisfy the condition \(f(x) = f(-x)\) for all \(x\) in its domain. Examples of even functions include \(f(x) = x^2\) and \(f(x) = \text{cos}(x)\).When a function is decomposed into an even and an odd part, the even part can be found using the formula:$$\text{Even part: } \frac{f(x) + f(-x)}{2}$$This formula effectively averages the values of the function at \(x\) and \(-x\), ensuring the result demonstrates the required symmetry across the y-axis.

odd function

An **odd function** has a different type of symmetry: it is reflected about the origin if you replace \(x\) with \(-x\) and then change the sign of the function. That means for a function \(f(x)\) to be odd, it must satisfy the condition \(f(x) = -f(-x)\) for all \(x\) in its domain. Examples of odd functions include \(f(x) = x^3\) and \(f(x) = \text{sin}(x)\). For decomposing functions, the odd part can be found using the formula:$$\text{Odd part: } \frac{f(x) - f(-x)}{2}$$This formula calculates the difference divisor between the function values at \(x\) and \(-x\), thus highlighting the function's symmetry about the origin.

function decomposition

The concept of **function decomposition** revolves around breaking down a complex function into simpler parts, making it easier to study and understand. In the context of even and odd functions, decomposition involves expressing a function as a sum of an even function and an odd function. This is especially useful as it allows you to analyze the symmetries and properties of each part individually. Let's consider \(f(x)\), which is neither even nor odd. We can write:\(f(x) = \text{Even part} + \text{Odd part}\)Using the formulas for even and odd parts mentioned above, we find:\begin{align*}\text{Even part: } & \frac{f(x) + f(-x)}{2} \text{Odd part: } & \frac{f(x) - f(-x)}{2}\rend{align*}The sum of these two parts will then reconstruct the original function.

symmetry in functions

Understanding **symmetry in functions** helps us grasp how certain transformations affect the graph of a function. There are several types of symmetry, but the most relevant ones here are:

• y-axis symmetry: Characteristic of even functions (\(f(x) = f(-x)\)).
  The function's graph looks the same on both sides of the y-axis.
• Origin symmetry: Characteristic of odd functions (\(f(x) = -f(-x)\)).
  The function's graph is reflected and inverted around the origin.

Recognizing these symmetries can simplify the process of recording, analyzing, and working with functions, especially when dealing with complex mathematical problems. The decomposition into even and odd functions essentially allows us to exploit these symmetries to better understand the function's behavior and properties.