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For each of the following functions, state whether it is even or odd or periodic. If periodic, what is its smallest period?

  1. $\sin ax \quad (a>0)$
  2. $e^{ax} \quad (a>0)$
  3. $x^m \quad (m= \text{ integer})$
  4. $\tan x^2$
  5. $|\sin (x/b)| \quad (b>0)$
  6. $x\cos ax \quad (a>0)$
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What are your thoughts? What did you try? Can you answer at least some of the questions for some of the functions? –  commenter Oct 26 '12 at 9:15
    
well for the first one, it is periodic since sin is periodic function, Likewise (5) is also. The rest are not. I am really stuck on the periodicity –  mary Oct 26 '12 at 9:21
    
That's correct: do you mean you can't prove your last comment and need hints on that? Or are you stuck on figuring out the smallest periods of (1) and (5)? Seeing if the functions are even or odd should be relatively easy: see if $f(-x) = f(x)$ or $f(-x) = -f(x)$. The first case is even and the second case is odd. –  commenter Oct 26 '12 at 9:31
    
It makes sense. Can you answer the question? I have some ideas, but they will be lacking some nice detail that I want to make sure I have. –  mary Oct 26 '12 at 9:33
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1 Answer 1

The definition of an even function is $f(-x) = f(x)$. The definition of an odd function is $f(-x) = -f(x)$. A periodic function means that for a fixed number $P$, $f(x + P) = f(x)$.

Therefore, substituting $-x$ in for $x$, $\sin(-ax) = -\sin(ax)$. This matches the definition of an odd function, so $\sin(ax)$ is odd.

To find if it is periodic, draw a graph of $\sin(ax)$ (I used WolframAlpha) and see if the graph repeats itself at all. In this case it does, every $2\pi$. Therefore the period is $2\pi$.

Try this method on the other functions and you should find it easy :)

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Thanks but there's some rigorous justification still missing unfortunately –  mary Oct 26 '12 at 9:41
    
I have corrected the $\LaTeX$ usage in your post; I will remark that mathematically it is incorrect, since the smallest period of $\sin(2x)$ would be $\pi$ and not $2\pi$; so the answer is actually $\frac{2\pi}a$. –  Asaf Karagila Oct 26 '12 at 10:11
    
@AsafKaragila : Thank you for the correction :) –  Jessica Oct 26 '12 at 10:16
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