My question is how can I show this? My problem states: Suppose m is an integer and f is the function defined by f(x) = xm.
Show that if m is an even number, then f is an even function.
If m is an even integer, then . Isn't that the very definition of an even function? That ...
If you need a more rigorous proof, since m is an even integer, you can write it as , where n is some other integer.
Huh? But that wouldn't justify whether the function equates to being even. I mean for example if we plug in some numbers such as 3 into the x. So it would look like 3^4*2. Yes, the n is even, BUT the function would equate to 6561. Sorry for my stupidity but please endure this with me.