Question

How do you determine if $f\left( x \right)=\dfrac{1}{{{x}^{2}}}$ an even or odd function is?

Answer 1

To find whether the given function is even or odd. We have to consider the given function as equation (1) and then by the formulas i.e. $\text{f}\left( x \right)$ is even if$\text{f}\left( -x \right)=\text{f}\left( x \right)$,$\text{f}\left( x \right)$ is odd if $\text{f}\left( -x \right)=-\text{f}\left( x \right)$. We have to check the given function and then we can say whether the function is even or odd by solving.

Complete step-by-step answer:
For the given problem we are given a function to determine whether it is an even or odd function.
For solving the above question let us consider the given function as equation (1).
$f\left( x \right)=\dfrac{1}{{{x}^{2}}}............\left( 1 \right)$
As we know the condition for proving whether the function is even or odd. i.e.
$\text{f}\left( x \right)$ is even if $\text{f}\left( -x \right)=\text{f}\left( x \right)$
$\text{f}\left( x \right)$ is odd if $\text{f}\left( -x \right)=-\text{f}\left( x \right)$
Let us consider
$\text{f}\left( x \right)$ is even if $\text{f}\left( -x \right)=\text{f}\left( x \right)..............case\left( 1 \right)$
$\text{f}\left( x \right)$ is odd if $\text{f}\left( -x \right)=-\text{f}\left( x \right)...........case\left( 2 \right)$
So, let us check our given function by doing$\text{f}\left( -x \right)$.
Calling the equation (1).
$f\left( x \right)=\dfrac{1}{{{x}^{2}}}$
Let us substitute $-x$in the equation (1) instead of x.
So, by substituting –x we get,
$\Rightarrow f\left( -x \right)=\dfrac{1}{{{\left( -x \right)}^{2}}}..........\left( 2 \right)$
Let us consider the above equation as equation (2)
By observing the above function we can see the term${{\left( -x \right)}^{2}}^{{}}$.
By squaring the term –x ‘-sign’ will be eliminated
Therefore, by rewriting the function we will get
$\Rightarrow f\left( -x \right)=\dfrac{1}{{{x}^{2}}}$
Let us consider the above equation as equation (3).
$\Rightarrow f\left( -x \right)=\dfrac{1}{{{x}^{2}}}..........\left( 3 \right)$
By comparing equation (3) with equation (1) we can see that RHS (Right hand side) of equation (3) is equal to RHS of equation (1). Therefore, we can say that
$\Rightarrow f\left( -x \right)=f\left( x \right)$
From case (1) we can say that given function is an even function.
Therefore, $f\left( x \right)=\dfrac{1}{{{x}^{2}}}$ is an even function.

Note: In the given function we can see only term ${{x}^{2}}$which is always an even function. We have to note a point that if we see any function raising the even power then it is an even function and also if we see any function raising the odd power then we can say that the given function is an odd function.