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

Oct 18, 2015

Always even (unless its domain is empty).

#### Explanation:

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Odd times odd

Suppose $f \left(x\right)$ and $g \left(x\right)$ are odd functions and $h \left(x\right) = f \left(x\right) g \left(x\right)$

By definition:

$f \left(- x\right) = - f \left(x\right)$ and $g \left(- x\right) = - g \left(x\right)$ for all $x$

So:

$h \left(- x\right) = f \left(- x\right) g \left(- x\right) = \left(- f \left(x\right)\right) \left(- g \left(x\right)\right) = f \left(x\right) g \left(x\right)$

$= h \left(x\right)$ for all $x$

So $h \left(x\right)$ is even.

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Even times even

Now suppose that $f \left(x\right)$ and $g \left(x\right)$ are even functions and $h \left(x\right) = f \left(x\right) g \left(x\right)$

By definition:

$f \left(- x\right) = f \left(x\right)$ and $g \left(- x\right) = g \left(x\right)$ for all $x$

So:

$h \left(- x\right) = f \left(- x\right) g \left(- x\right) = f \left(x\right) g \left(x\right) = h \left(x\right)$ for all $x$

So $h \left(x\right)$ is even.

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Exception

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

For example:

Let $f \left(x\right) = {\cos}^{- 1} \left(x\right)$ and $g \left(x\right) = {\sec}^{- 1} \left(\frac{x}{2}\right)$.

Then the domain of $f \left(x\right)$ is $\left[- 1 , 1\right]$ and the domain of $g \left(x\right)$ is $\left(- \infty , - 2\right] \cup \left[2 , \infty\right)$. They are both even functions.

The domain of $f \left(x\right) g \left(x\right)$ is $\left[- 1 , 1\right] \cap \left(\left(- \infty , 2\right] \cup \left[2 , \infty\right)\right) = \emptyset$