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About LinkBacksMy 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
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If you need a more rigorous proof, since m is an even integer, you can write it as, where n is some other integer.
![\displaystyle \begin{align*} (-x)^m &= (-x)^{2n} \\ &= \left[(-x)^2\right]^n \\ &= \left(x^2\right)^n \\ &= x^{2n} \\ &= x^m \end{align*}](http://latex.codecogs.com/png.latex?\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 thatOriginally Posted by Eraser147
for all x.
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