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Is there a way to see if a relation is a function without having to do a "vertical line test" (where you draw a vertical line on the graph and if there line touches two points then it's not a function).

To determine if a function is even or odd you simply go f(x) = f(-x); even, f(-x) = -f(x); odd. Can I do something similar to find out if a relation is a function?

Thanks

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  • $\begingroup$ Your example of even and odd functions isn't really an algebraic way of solving. $\endgroup$ – Simply Beautiful Art Feb 17 '16 at 23:06
  • $\begingroup$ Ok, what type of method is it? $\endgroup$ – frog1944 Feb 17 '16 at 23:09
  • $\begingroup$ I don't know. But that's definitely not 'solved' algebraically, I can tell you that much. $\endgroup$ – Simply Beautiful Art Feb 17 '16 at 23:10
  • $\begingroup$ Ok, thanks @Simple Art. $\endgroup$ – frog1944 Feb 17 '16 at 23:12
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For a relation to be a function, it must be one-to-one or injective, meaning that it must map each input into a different output.

If you can't use the vertical line test, see if you can determine whether or not the function/relation has branches.

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  • $\begingroup$ Ok, can you do that algebraically? Like for even and odd functions I can go; if f(x) = f(-x) it's even and if f(-x) = -f(x). Can I do something similar to see if a relation is a function? $\endgroup$ – frog1944 Feb 17 '16 at 22:58
  • $\begingroup$ @frog1944 I would look at what operations are in the function and determine if any of them have branches. i.e. logs, trig, roots, exponentials, etc. $\endgroup$ – Simply Beautiful Art Feb 17 '16 at 23:00
  • $\begingroup$ Ok, so there is no similar way to find out if it's a function other than looking at the function and determining if it has any branches? $\endgroup$ – frog1944 Feb 17 '16 at 23:08
  • $\begingroup$ @frog1944 Well, you can break it up into parts and pieces if that makes it easier. Also look to see if it is defined for all $x$, else it isn't a relation. And of course, this depends on if you consider $x\in\mathbb{R}$ or something different. But no, not that I think. $\endgroup$ – Simply Beautiful Art Feb 17 '16 at 23:11
  • $\begingroup$ Awesome! Thanks :) $\endgroup$ – frog1944 Feb 17 '16 at 23:12

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