Parity of function (simple question)
Is correct to ask if $f(x)={\sqrt{1-x^{2}}}$ is an even or odd function? How to define the parity of a function?
$\endgroup$ 21 Answer
$\begingroup$A function $f$ is defined to be even if $f(-x) = f(x)$ for all $x$ such that $f(x)$ is actually defined. It is odd if $f(-x) = -f(x)$ for all $x$.
In this case, we have
$$f(-x) = \sqrt{1 - (-x)^2} = \sqrt{1 - x^2} = f(x)$$
$\endgroup$ 1
