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Given is a function f(x)= 1-x with 0 < x < 1 that should be extended on the interval [-1, 1] to get an odd function. Outside this interval it should be continued with a period of 2.

If it would be an even function, I would extend it to f(x) = 1-|x| as even functions are symmetric around the y-axis. Would this be correct?

Now I am not sure, how to extend it to an odd function. I remember that odd functions should be point symmetric around the origin, and I think that in this case it will stay the same: f(x)= 1-x. Can this be correct?

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You need $1-x$ on $(0,\,1)$ so on $(-1,\,0)$ the result is $-1-x$. There is a discontinuity at $0$, where any odd function must be $0$. By periodicity, the function should vanish at all even integers. You can work out the function elsewhere; note that, except at discontinuities, the gradient is always $-1$.

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  • $\begingroup$ Thanks a lot. This really makes sense. :) $\endgroup$ – mrs fourier Apr 22 '17 at 16:28

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