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An odd function is symmetrical in the 1st and 3rd quadrants. Does this means that it always passes through the origin?

Jonas Meyer
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user168467
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3 Answers3

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As André Nicolas showed, under your conditions and if $f(0)$ exists, $f(0)=0$. However, nothing in your question implies that $f(0)$ must exist.

If you let $f(x)=\frac1x$ then $f$ is a symmetrical odd function, its graph is in quadrants I and III, but $f(0)$ is undefined.

So, you can say "$f(0)$ is either $0$ or undefined." Or, if you want to stick to terminology about graphs, "the graph of $f$ either passes through the origin or it does not intersect the $y$-axis at all."

Rory Daulton
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    I think your conclusion is too strong. I have in mind some variation on $\sin \frac{1}{x}$ which is odd, discontinuous, and passes $0$ infinitely often. – nomen Aug 10 '14 at 03:34
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    @nomen: But $\sin \frac{1}{x}$ never intersects the $y$ axis. (It's undefined at $x=0$.) The statement holds. And if you have an equation where $y$ has multiple values at $x=0$, what you have by definition isn't a function of $x$. – cHao Aug 10 '14 at 04:41
  • You're right, I misunderstood. – nomen Aug 10 '14 at 04:57
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Let $f(x)$ be odd, and defined at $x=0$. Then $f(-x)=-f(x)$ for all $x$. In particular, $f(-0)=-f(0)$, so $f(0)=-f(0)$. It follows that $f(0)=0$. The "curve" $y=f(x)$ passes through the origin.

André Nicolas
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    Well, it follows that the function equals 0 at the origin, but to say it "passes" through the origin is misleading language in my opinion. What if $f(x)=\operatorname{sgn}x$ for $x\neq 0$? – David H Aug 09 '14 at 16:03
  • @user168467: You are welcome. Many questions about odd or even functions can be answered by a semi-mechanical use of the definition. – André Nicolas Aug 09 '14 at 16:04
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    What if the function is not defined around the origin? It can still be odd can't it? – Klaas van Aarsen Aug 09 '14 at 16:16
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    @IlikeSerena: It certainly can be. Often unless a specification to the contrary is made, it is assumed that a function is defined everywhere. Thank you for your comment, I have made the requirement that $f$ is defined at $0$ explicit. – André Nicolas Aug 09 '14 at 16:21
  • I hope you don't mind me nitpicking. It seems to me that for the function to pass through the origin, it should be defined in an interval around the origin. – Klaas van Aarsen Aug 09 '14 at 16:26
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    I certainly don't mind the "nitpicking," which I would rather call being precise, and which is part of our DNA as mathematicians. "Passes through the origin" is an informal term whose formal definition, if one were given, perhaps should include the condition of being defined in an interval, and should perhaps include some smoothness condition. But I have not seen a formal definition of "passes through $P$," and took the broadest possible interpretation. – André Nicolas Aug 09 '14 at 16:35
  • This answer is wrong. An off function do not need to be defined for $x=0$. – kjetil b halvorsen Aug 09 '14 at 18:20
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    The answer explicitly assumes that $f$ is defined at $0$. So probably "wrong" is not quite right, though "incomplete" certainly would be. – André Nicolas Aug 09 '14 at 18:32
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If any odd function is defined at $x=0$ then it must pass through origin other wise it can remain symmetrical about origin without actually passing through it.

Mohsen Shahriari
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SINGH
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