When you multiply two odd or two even functions, what type of function will you get?

1 Answer
Oct 18, 2015

Answer:

Always even (unless its domain is empty).

Explanation:

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Odd times odd

Suppose #f(x)# and #g(x)# are odd functions and #h(x) = f(x)g(x)#

By definition:

#f(-x) = -f(x)# and #g(-x) = -g(x)# for all #x#

So:

#h(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x)#

#= h(x)# for all #x#

So #h(x)# is even.

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Even times even

Now suppose that #f(x)# and #g(x)# are even functions and #h(x) = f(x)g(x)#

By definition:

#f(-x) = f(x)# and #g(-x) = g(x)# for all #x#

So:

#h(-x) = f(-x)g(-x) = f(x)g(x) = h(x)# for all #x#

So #h(x)# is even.

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Exception

If the domains of #f(x)# and #g(x)# do not intersect, then their product #f(x)g(x)# has an empty domain, so is the empty function. The empty function probably does not count as odd or even.

For example:

Let #f(x) = cos^(-1)(x)# and #g(x) = sec^(-1)(x/2)#.

Then the domain of #f(x)# is #[-1, 1]# and the domain of #g(x)# is #(-oo, -2] uu [2, oo)#. They are both even functions.

The domain of #f(x)g(x)# is #[-1, 1] nn ((-oo, 2] uu [2, oo)) = O/#