Question

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

Answer 1

A function $f\left( x \right)$ is said to be an even function if it satisfies the following condition: $f\left( -x \right)=f\left( x \right).$ A function $f\left( x \right)$ is said to be an odd function if it satisfies the following condition: $f\left( -x \right)=-f\left( x \right).$ We verify if either of these conditions is satisfied by the given function.

Complete step by step answer:
Let us consider the given function $f\left( x \right)={{x}^{2}}-x.$
We are asked to check if the given function is an even or an odd function.
To check if this function is an even function, we need to apply the condition for an even function. That is, $f\left( -x \right)=-f\left( x \right).$
Consider the function, $f\left( x \right)={{x}^{2}}-x.$
When we apply the condition for even functions, we replace all $x$ in the equation with $-x.$
We get,
$\Rightarrow f\left( -x \right)={{\left( -x \right)}^{2}}-\left( -x \right)$
And from this we get,
$\Rightarrow f\left( -x \right)={{x}^{2}}+x$
Since, ${{\left( -a \right)}^{2}}={{a}^{2}}$ and $-\left( -a \right)=+a.$
And from this we can see that $f\left( -x \right)={{x}^{2}}+x\ne {{x}^{2}}-x=f\left( x \right).$
That can be written as $f\left( -x \right)\ne f\left( x \right).$
Since the given function does not hold the property of an even function, we conclude that the given function is not an even function.
Now, there is a possibility that the given function is an odd function.
Before confirming that the given function is an odd function, let us check if the given function holds the property of an odd function.
Since we have $f\left( -x \right)={{x}^{2}}+x,$ what remains to find is the value of $-f\left( x \right).$
So, we get,
$\Rightarrow -f\left( x \right)=-\left( {{x}^{2}}-x \right)$
When we open the brackets, we get
$\Rightarrow -f\left( x \right)=-{{x}^{2}}+x$
And from this we can see that
$\Rightarrow f\left( -x \right)={{x}^{2}}+x\ne -{{x}^{2}}+x=f\left( x \right)$
That means,
$\Rightarrow f\left( -x \right)\ne -f\left( x \right)$
This implies that the given function does not hold the property of an odd function also.
That implies, the given function $f\left( x \right)={{x}^{2}}-x$ is neither an even function nor an odd function.

Note:
Unlike real numbers, there are functions which are neither even nor odd. You should not confirm that a function is odd without checking if it satisfies the property of odd function when you find it does not satisfy the property of even function. Similarly, do not make a conclusion that a function is even before checking the property of even functions.