# How to show a function is even/odd

by mcfc
Tags: even or odd, function
 P: 18 1. The problem statement, all variables and given/known data Hi, I'm new to this site, I've had a look around and there are alot of useful sections, particularly the section with math and science learning materials. Anyway, I need to show that the following function is odd $$f(x)=\left\{\begin{array}{ccc} -\sin x&\mbox{ for }-\pi \leq x< \frac{-\pi}{ 2}\\ \sin x &\mbox{ for } \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}\\ -\sin x &\mbox{ for } \frac{\pi}{2}  P: 200 If f(-x)=f(x), then the function is even. If f(-x)=-f(x), then the function is odd.  Sci Advisor HW Helper PF Gold P: 4,097 This might be a useful variation of the definitions: If the function f(x) is even, then f(x)-f(-x)=0 for all x. If the function f(x) is odd, then f(x)+f(-x)=0 for all x. P: 18 ## How to show a function is even/odd  Quote by robphy This might be a useful variation of the definitions: If the function f(x) is even, then f(x)-f(-x)=0 for all x. If the function f(x) is odd, then f(x)+f(-x)=0 for all x. To show it's odd: look at values in the intervals? [tex]-\sin(-\pi) - \sin({\pi }) = 0$$

$$\sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0$$

$$-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0$$

do I need to show anything else?
P: 32
 Quote by mcfc To show it's odd: look at values in the intervals? $$-\sin(-\pi) - \sin({\pi }) = 0$$ $$\sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0$$ $$-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0$$ do I need to show anything else?
You would have to show that it's true for every value in the interval, not just at a few random points. So you'd have to let $$a$$ be a random value in each interval, and then look at $$f(a)$$ and $$f(-a)$$. Since the intervals are symmetric, once you've assigned an interval for $$a$$, it will be obvious what interval $$-a$$ is in and therefore which definition of the function you need to use.

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