91Ó°ÊÓ

In precalculus, determining whether a function is odd, even, or neither is crucial for understanding its symmetry and behavior. To determine a function's parity, we test it against two key conditions:

• A function is even if it satisfies the condition \(f(-x) = f(x)\).
• A function is odd if it satisfies the condition \(f(-x) = -f(x)\).

Understanding these conditions helps us categorize the function correctly. If a function does not satisfy either of these conditions, it is neither odd nor even. Let's apply these concepts to a specific function: \(f(x)=x \cos x\).

Even functions have the distinct property of being symmetrical about the y-axis. This means that the value of the function remains the same whether you substitute \(x\) or \(-x\). Mathematically, this is written as:\br\[f(x) = f(-x)\]\brTo test if a function is even, you substitute \(-x\) for \(x\) and see if you get back the original function. Let's illustrate with an example:

Given a function \ f(x) = x^2 \, calculate \ f(-x) :
\ \ f(-x) = (-x)^2 = x^2\
Since \ f(-x) = f(x) \, this function is even.
In our given exercise with \ f(x) = x \cos(x) \, substituting \ (-x) \ gives us \ f(-x) = -x \cos(x) \. Since this is not equal to our original function, \ f(x) = x \cos(x) \, we conclude this function is not even.

Odd functions possess a type of symmetry called origin symmetry. For a function to be odd, substituting \( -x \) for \( x \) should result in the negative of the original function. Formally, this can be stated as:

\[f(-x) = -f(x)\]
To test if a function is odd, substitute \(-x\) and check if you obtain \ -f(x) \.
In the given exercise, we had \ f(x) = x \cos(x) \. Substituting \(-x\) we got:
\ \ f(-x) = -x \cos(x) \ which simplifies to \f(-x) = - x \cos(x)\.
This satisfies the condition for being an odd function because it matches \-f(x)\. Therefore, \ f(x) = x \cos(x)\text{ is an odd function.}\

Function substitution involves replacing the variable \(x\) in a function with another expression. It is a key tool for determining the parity of functions. Let's delve deeper into how we use this technique while working with our exercise.

When we examined the given function \ f(x) = x \cos(x) \:

• Step 1: First, we replaced \(x\) with \(-x\). This gave us \ f(-x) = -x \cos(-x) \.
• Step 2: We know that \ cos(-x) = cos(x) \. So, we simplified it to \ f(-x) = -x \cos(x) \.

Using substitution, we compared \ f(x) \ and \ f(-x) \. We noticed that \ f(-x) = -f(x)\, signaling that the function is odd. This step-by-step substitution is an effective strategy in testing the functionality for parity.