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Question: How do you determine if \[f\left( x \right) = \left| {{x^2} + x} \right|\] is an even or odd functio...

How do you determine if f(x)=∣x2+x∣f\left( x \right) = \left| {{x^2} + x} \right| is an even or odd function?

Explanation

Solution

A function can be defined as even or odd if it satisfies an even or odd definition. In the given function, we need to determine algebraically whether a function is even or odd. Hence, to determine if f(x)=∣x2+x∣f\left( x \right) = \left| {{x^2} + x} \right| is an even or odd function, we need to consider the following:
If f(x)=f(−x)f\left( x \right) = f\left( { - x} \right), then f(x)f\left( x \right) is even: Even functions have symmetry about the y-axis.
If f(−x)=−f(x)f\left( { - x} \right) = - f\left( x \right), then f(x)f\left( x \right) is odd: Odd functions have symmetry about the origin.

Complete step by step answer:
Given,
f(x)=∣x2+x∣f\left( x \right) = \left| {{x^2} + x} \right|
Let us determine for even function:
We know that for even: f(x)=f(−x)f\left( x \right) = f\left( { - x} \right), hence:
f(−x)=(−x)2−(−x)f\left( { - x} \right) = {\left( { - x} \right)^2} - \left( { - x} \right)
Simplifying the terms, we get:
⇒f(−x)=x2+x≠f(x)\Rightarrow f\left( { - x} \right) = {x^2} + x \ne f\left( x \right)
Since, f(x)≠f(−x)f\left( x \right) \ne f\left( { - x} \right), then the given function f(x)f\left( x \right) is not even.
Now, let us determine for odd function:
We know that for odd: f(−x)=−f(x)f\left( { - x} \right) = - f\left( x \right), hence:
−f(x)=−(x2−x)- f\left( x \right) = - \left( {{x^2} - x} \right)
Simplifying the terms, we get:
⇒−f(x)=−x2+x≠f(−x)\Rightarrow - f\left( x \right) = - {x^2} + x \ne f\left( { - x} \right)
Since, f(−x)≠−f(x)f\left( { - x} \right) \ne - f\left( x \right), then the given function f(x)f\left( x \right) is not odd.
Thus, the given function f(x)=∣x2+x∣f\left( x \right) = \left| {{x^2} + x} \right| is neither odd nor even.

Note: The key point to note is that, we must know the conditions, if f(x)=f(−x)f\left( x \right) = f\left( { - x} \right), then f(x)f\left( x \right) is even and iff(−x)=−f(x)f\left( { - x} \right) = - f\left( x \right), then f(x)f\left( x \right) is odd. A function can be neither even nor odd if it does not exhibit either symmetry. Also, the only function that is both even and odd is the constant function. A function ‘f’ is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(−x)=f(x)f\left( { - x} \right) = f\left( x \right) for all x in the domain of f. A function f is odd if the graph of f is symmetric with respect to the origin.