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# How do you determine if $f(x) = 1$ is an even or odd function?

Last updated date: 09th Aug 2024
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Hint: A function $f(x)$ is said to be an even function when its output at $x\;{\text{and}}\;( -x)$ for all value of $x$ is equal. Mathematically it can be understood as follows $f(x) = f( - x),$ for all $x$ belongs to the domain of the function. And a function $f(x)$ is said to be an odd function when the sum of outputs at $x\;{\text{and}}\;( - x)$ for all values of $x$ is equal to zero. Mathematically it can be understood as follows $f(x) + f( - x) = 0,$ for all $x$ belongs to the domain of the function.

Complete step by step solution:
In order to determine whether the given function $f(x) = 1$ is an even or odd function, we will find it by putting the same argument in the function with positive and negative signs separately then we will figure out whether it is even or odd.
Taking the argument equals to $a$
$\Rightarrow f(a) = 1$
Now taking argument equals to $( - a)$
$\Rightarrow f( - a) = 1$
That gives the same value because this is a constant function and constant functions do not get affected by their arguments.
From above we get, $f(a) = f( - a)$
And $g(x)$ is an even or odd function when $g(x) = g( - x)\;{\text{or}}\;g(x) = - g( - x)$ respectively.
$\therefore f(x) = 1$ is an even function.

Additional information: You can also say about a function whether it is even or odd with help of its graph, if its graph has symmetry about x-axis then it is even and if it has symmetry about the origin then it is odd

Note: All constant functions except $f(x) = 0$, are even functions because their output don’t get affected by any argument. $f(x) = 0$ is a special case hence it is both odd and even function. And if a function is given as $f(x) = {x^n}$, then it will be even or odd depending if $n$ is either even or odd.