Is it possible that function is even as well as decreasing in its domain?
Of course, let the domain be
and let the function be
Since the definition of an even function is that
, can you have an even function whose domain is on only one side of zero?
The definition is
, it just so happens that my set of
is also in the domain is empty, so since we have a universal quantifier applied to an empty set, it is vacuously true that my function is even.
Good evening sir
sorry sir but one of my great sir has given this explanation (which confused me more )as he disagree with your explanation .His explanation is following
Any of the common definitions for even functions
A function such that
in the domain of
A function having its graph symmetrical with respect to the
in fact implies that the domain
of an even function
is itself symmetrical with respect to the origin, i.e.
. This is for example how the notion of even function is introduced in Romanian textbooks: a function
Thus speaking about an even function defined on
is disallowed, and not vacuously true (only because
does not exist).
I've never seen that assumption made, before. If that is your definition of an even function, then you may ignore what I've said, but I don't see any reason why one should assume the domain is symmetric and I never have. It's easy to *extend* even functions to a maximal, symmetric domain, but there's no reason do do that ahead of time IMO.
In that case you need to see if weakly decreasing is possible, if so, then the identically zero function works, otherwise the answer is no for the reason explained above (I'm assuming the function is
because this is a calculus question, you can get it for continuous functions without too much work as well.