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) = x^{m. }
Show that if m is an even number, then f is an even function.
If m is an even integer, then $\displaystyle \displaystyle \begin{align*} (-x)^m = x^m \end{align*}$. Isn't that the very definition of an even function? That $\displaystyle \displaystyle \begin{align*} f(-x) = f(x) \end{align*}$...
If you need a more rigorous proof, since m is an even integer, you can write it as $\displaystyle \displaystyle \begin{align*} x^{2n} \end{align*}$, where n is some other integer.
$\displaystyle \displaystyle \begin{align*} (-x)^m &= (-x)^{2n} \\ &= \left[(-x)^2\right]^n \\ &= \left(x^2\right)^n \\ &= x^{2n} \\ &= x^m \end{align*}$
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.
An even function is NOT a function which only produces even numbers. An even function is a function that is a complete reflection of itself in the y - axis (in other words, inputting a negative number will give you the same answer as a positive one). In symbols, an even function has the property that $\displaystyle \displaystyle \begin{align*} f(-x) = f(x) \end{align*}$ for all x.