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I'm in a calc I class where I'm faced with the question:

Suppose that f(x) is an odd function, and periodic with period 10. If f(3) = 4, find f(7) + f(5).

Unfortunately, this is not talked about in our text, which says to me I already have this knowledge but can't seem to make sense of it.

I know that an odd function has the property f(-x) = -f(x), but how does that help if I don't know f(x)?

What I was able to find thus far has been tied to more complex ideas which I don't know and we have yet to cover. It said a function is periodic if f(x + T) = f(x). Is this going in the right direction, and if so how can go about dumbing it down a tad?

Any links to info, hints towards a direction, an explanation of the type of problem this is would be appreciated.

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Hint: For what $x$ does $f(x)=f(x+10)=f(7)?$ For what $x$ does $f(x)=f(x+10)=f(5)?$ Once you have written these out, use the oddness and periodicity to find $f(7)$ and $f(5)$ - it should be quite straightforward.

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    I saw that. Yes, you make it plenty clear about -5, but I feel like I have to make assumptions about f(x) to be able to solve for it. Like with sin: sin pi/2 = 1 & sin 3pi/2 = -1 & sin pi = 0. f(5) is equal units between f(3) and f(7), but can assume 0?2012-09-26
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    @RobertoWilko You're going off-route. You got $f(-5)=f(5)$ right? What is $f(-5)$ also equal to?2012-09-26
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    Yep, this is the problem. I get into a never ending loop of a value I don't know, or don't see a way to come up with unless I chop up the period.2012-09-26
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    @RobertoWilko I can't make it more clear: you have $f(5)=-f(5)$. What is the *only* number that is both positive and negative?2012-09-26
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    0? If so, that's what I was eluding to with sin? If not, and I'm a lost cause for you, I'll understand!2012-09-26
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    @RobertoWilko Yes $0$. I don't see how you said about sin applies here, but from what you've wrote, be careful: if a function has a period $2t$, and, say $f(0)=-1,\;f(t)=1$ this does *not* mean $f(t/2)=0$. The fact that the variable is "in the middle" does not mean the output will be "in the middle". Hope this helps2012-09-26
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    I was sure it didn't mean that. In fact, as usual, I was overlooking the most obvious point you made. I sat here saying they equal each other forever and neglected to realize exactly what I was saying. Thanks so much for your time and patience!2012-09-26
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Hint: You correctly translated "odd function". You know $f(3)$, so you know $f(-3)$ and similarly for any couple of opposite values of the argument of $f$.
Now, $f$ is periodic of period $10$, so $f(3)=f(3-10)=f(-7)$. But then...

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    f(-7) = -f(7) = -4? It makes sense how you use the period to determine another x value. I'm struggling with 5,-5! I'm sorry, I've recently been put on meds which make it difficult for me to focus.2012-09-25
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    $f(3)=f(-7)=4 \longrightarrow f(7)=-f(-7)=-4$. About the rest of the exercise, as Julien said, surely $f(5)=f(-5)$ for periodicity. But also $f(5)=-f(-5)$ because the function is odd. If $f(5)$ equals a number *and* its opposite, then it's $0$.2012-09-26