Rating

We will use $f(x) = x^2\sin{x}$ (defined for all real numbers) as an example.

For a function to be considered for evenness or oddness, we require that its domain of definition (where the function is defined) is symmetric around the origin.

For instance, it does not make sense to ask whether $y = \sqrt{x}$ is an odd function or not, because it is only defined on the set of non-negative numbers $[0,+\infty) = \{x\in\mathbb{R}: x\ge 0\}$.

For our example $f(x) = x^2\sin{x}$, we have

$f(-x) = (-x)^2\sin(-x) = x^2(-\sin{x}) = -x^2\sin{x}. $

If $f(-x) = f(x)$ holds for all $x$ in the domain of definition, then the function is even.

If $f(-x) = -f(x)$ holds for all $x$ in the domain of definition, then the function is odd.

For our example $f(x) = x^2\sin{x}$, we computed $f(-x) = -x^2\sin{x} = -f(x)$, so we know that it is odd.

If neither is true, then the function is neither even nor odd. For instance $f(x) = x + 1$ is neither even nor odd, because $f(1) = 2$ and $f(-1) = 0$, so for $x = 1$, neither of the above conditions holds.

It is in fact possible for a function to be both odd and even. However, this is possible only when a function has a domain of definition that is symmetric around the origin and is constantly equal to $0$ whenever defined. Because being even means $f(-x) = f(x)$, and being odd means $f(-x) = -f(x)$, and these two relations tell us $f(x) = -f(x)$, which implies $f(x) = 0$.

Consider $f(x) = \ln{|\frac{1-x}{1+x}|}$ which is defined for $x\not= -1, 1$.

We compute $f(-x) = \ln{|\frac{1+x}{1-x}|} = -\ln{|\frac{1-x}{1+x}|} = -f(x)$. So this function is odd, although it is not obvious from the way we write it.

If you multiply two even or two odd functions together, you would get an even function. For example, $f(x) = x^2$ and $g(x) = \cos{x}$ are both even, so $f(x)g(x) = x^2\cos{x}$ is also even.

If you multiply an odd function and an even function together, you get an odd function. For instance, $f(x) = x^3$ is odd and $g(x) = \cos{x}$ is even, so $f(x)g(x) = x^3\cos{x}$ is odd.

If a function is obtained by composing a number of even or odd function, the result is even unless all of them are odd. For instance, $f(x) = x^3$ is odd, and so is $g(x) = \sin{x}$, so both $f(g(x)) = (\sin{x})^3$ and $g(f(x)) = \sin(x^3)$ are odd. However, $\cos(x^3)$ and $(\cos{x})^3$ are even, because $\cos{x}$ is even and $x^3$ is odd.

An odd function $f$ which has a definition at $x = 0$ must pass through the origin, because by putting $x = 0$ in the condition for odd function, we get $f(-0) =-f(0)$, so $f(0) = 0$.

As a consequence, if we know $f(0)$ is not $0$, we can say with certainty that the function is not odd.