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.


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.

  • $\begingroup$ 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? $\endgroup$ – ChiefTwoPencils Sep 26 '12 at 0:34
  • $\begingroup$ @RobertoWilko You're going off-route. You got $f(-5)=f(5)$ right? What is $f(-5)$ also equal to? $\endgroup$ – user39572 Sep 26 '12 at 0:36
  • $\begingroup$ 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. $\endgroup$ – ChiefTwoPencils Sep 26 '12 at 0:44
  • $\begingroup$ @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? $\endgroup$ – user39572 Sep 26 '12 at 0:47
  • $\begingroup$ 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! $\endgroup$ – ChiefTwoPencils Sep 26 '12 at 0:50

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...

  • $\begingroup$ 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. $\endgroup$ – ChiefTwoPencils Sep 25 '12 at 23:55
  • $\begingroup$ $f(3)=f(-7)=4 \longrightarrow f(7)=-f(-7)=-4$. 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 $0$. $\endgroup$ – Andrea Orta Sep 26 '12 at 15:32

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