Question

How do you determine if $xy = 1$ is an even or odd function?

Answer 1

Let’s assume that $y$ is a function of $x$ .
We will put $- x$ in the equation and if it gives us $f\left( x \right)$ , it is an even function. On the other hand if it gives us $- f\left( x \right)$ , it is an odd function.

Complete step-by-step solution:
Assuming that $y$ is a function of $x$ , so,
$y(x) = \dfrac{1}{x}$
This function is odd. Example let, $x = 5$ . Then the x value is less than the value of $1$ .
Now, $x$ value substituting in the equation, $y(5) = \left( {\dfrac{1}{5}} \right)$ . The value is $0.2$ .The value is nearest the odd number and it is less than the value of $1$ .
To determine whether the function is even or odd, we evaluate $y( - x)$ in terms of $y\left( x \right)$ ,
Now put $x = - x$ , substituting in the equation,
$y( - x) = \dfrac{1}{{ - x}} = - \left( {\dfrac{1}{x}} \right) = - y(x)$ from the condition,
We can say that the given function is an odd function.
The even function is, when putting $x = - x$ gives us $y\left( x \right)$ only.
For example: Let $y(x) = {x^2}$
$y( - x) = {( - x)^2} = {x^2} = y(x)$ , from the condition
Now, put $x = 3$ , $y(3) = {(3)^2}$ . The value is 9. Since we got our function as similar to the function we assumed, our assumed function is even.

The given function is an odd function.

Note: Some functions, unlike integers, can be both odd and even. For example: $y(x) = 0$ and $y(x) = x + 1$ . A function is even if F of negative is equal to f of $x$ . So, if you replace $x$ with negative $x$ and there is no change, the new function that you get looks exactly like the original function and then it is even. Now what about if it’s odd if F of negative $x$ is equal to negative f of $x$ . So, if you replace negative $x$ which acts everything in the function.