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So, if you test the function for symmetry and it turns that it has one of these, does it mean that it cannot have the others? Therefore, one function has

**one**type of symmetry

**only**?

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So, if you test the function for symmetry and it turns that it has one of these, does it mean that it cannot have the others? Therefore, one function has

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AlephZero

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No. What about a circle, x^2 + y^2 = a^2?Therefore, one function hasonetype of symmetryonly?

(I assume you are talking about "multi-valued functions", as in the usual "sketch the shape of the graph" type of question, since you said "symmetry about the y axis".)

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A full circle dosen't fall within the definition of a function. To answer the question, the only function symmetrical around the x axis is f(x) = 0. And no, a continuous function cannot be both symmetrical around the y axis and the origin, since that would imply two outputs for one input on a certain interval.

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I think you'll find you're doing a major injustice to, at the very least, group theorists with that statement! :surprisedI know there are 3 kinds of symmetry: x-axis, y-axis, and origin.

See http://mathworld.wolfram.com/Symmetry.html and links within for just the tip of the symmetry iceberg!

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mathwonk

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e.g. the group taking (x,y) to itself or to (-x,y) gives what you call y axis symmetry. the group taking (x,y) to itself or to (-x,-y) I gues gives origin symmetry, etc....but there are many other more complicated groups.

e.g. the circle (rotation) group acts on the plane, and leaves x^2 + y^2 invariant, as mentioned above

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Ok, I suppose you are only asking about real functions of one variable,like f:R->R. You only have to think of how symmetry is "translated" in algebra.

So, if you test the function for symmetry and it turns that it has one of these, does it mean that it cannot have the others? Therefore, one function hasonetype of symmetryonly?

If a function has a x-axis summetry, then it simply is not a function, with the classic definition, or it's a multivalued function, because the point f(x) must have 2 values, y and -y, in order for the function to be x-symmetric.

If a function has y-axis summetry, then it can be a "normal" function, as this symmetry means that f(x)=f(-x) for any x in the domain of the function.

Finally, if it is symmetric around (0,0) then this condition must be true: f(-x)=-f(x) .

So you can now check for yourself, with the help of these equations, if 2 of the above conditions can be true "simultaneously" (ok sorry for my bad english...)

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AlephZero

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Define f(t in [0, 2pi) -> (x,y) in R^2 : x = a cos t, y = a sin t)A full circle dosen't fall within the definition of a function.

Then f is a function which defines a full circle, I think.

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What those guys who were referring to Group Theory were saying is that are many, many more symmetries than just the three you named, but that's okay - you don't need to know about the others, just so long as you understand these three.

I think COnfused gave the best direct answer to your question. Did you understand what he was getting at, or do you still need some more help?

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That's not 100% right, though most relations that have x-axis symmetry are indeed not functions.If a function has a x-axis summetry, then it simply is not a function, with the classic definition, or it's a multivalued function, because the point f(x) must have 2 values, y and -y, in order for the function to be x-symmetric.

There are indeed lots of types of symmetry but one other that is common in high school is symmetry about the line y=x. If (x,y) is on the graph of a relation if and only if (y,x) is on the graph, for all x and y, then it is symmetric about the line y=x. For functions, I think it would be symmetric about y=x if f(f(x))=x for all x in the domain of f. Not totally sure on that...examples of functions with this symmetry are 1/x, -1/x, -x, -x+b for any b, (1-dx)/(d+x) for any d. Ah yes, the test f(f(x))=x is false because f(x)=0 for all x has this property but is not symmetric about y=x.

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