How do we determine whether a function *f* is even, odd or not
symmetric at all? There is a very simple algorithm using the definition.

**1.** First we look at the domain
*D*( *f* ).*f*
cannot be symmetric either. For instance, the function
*f* (*x*) = ln(*x* + 3)*D*( *f* ) = (−3,∞),*f* cannot be symmetric. On the other hand, the function
*g*(*x*) = arcsin(*x*^{2}) + 1/*x**D*(*g*) = [−1,0) ∪ (0,1],

**2.** If the domain is a symmetric set, we do the main test. We
substitute *x**f* and try to get rid of the
sign. If we can make this extra sign disappear, that is, if we obtain
*f* (*x*)*f* (−*x*),*f* is
even. On the other hand, if we are able to pull this sign out so that we get
*f* (*x*)*f* (−*x*),*f* is
odd. If none of this is possible, then the function is not symmetric. For
making the sign disappear or jump out we use our knowledge of
elementary functions.

Note that although usually it is easy to see whether the function is
symmetric or not, sometimes it is not so simple. In particular, sometimes it
happens that you substitute *x**f* (*x*)*f* (*x*),*f* (−*x*) = *f* (*x*),*f* and try to see whether this is true for all
*x* or not. If yes, the function is even, if not (if it fails for even
one *x* from the domain), then you can be sure that it is not even.
Similarly you try to solve
*f* (−*x*) = −*f* (*x*),*x*, then the function is surely not odd.

**Example:** Determine the symmetry of

**Solution:**
As we saw above, the domain is a symmetric set, so the function might be
symmetric. We substitute -*x*:

We used the fact that *x*^{2}*x*)^{2} = *x*^{2}.*f* (*x*)*f* (*x*).

In both cases we obtained an equation that is not true for all *x* from
the domain. We conclude that the function is not symmetric.

Note that in the arcsin part, the sign disappeared. This shows that the
function *x*^{2})*x*.*x*

For further examples we refer to Solved Problems.

Periodicity

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properties of real functions