If $f$ and $g$ are both even functions, is the product $fg$ even? If $f$ and $g$ are both odd functions, is $fg$ odd? What if $f$ is even and $g$ is odd? Justify your answers.

Key Concepts: Even Function, Odd Function, Product Of Functions Explanation: An even function is a function that satisfies $f(-x)=f(x)$ for all $x$ in its domain. An odd function is a function that satisfies $f(-x)=-f(x)$ for all $x$ in its domain. Let $f$ and $g$ be even functions. Then, $f(-x)=f(x)$ and $g(-x)=g(x)$ for all $x$ in the domain of $f$ and $g$. So we have $(fg)(-x)=f(-x)g(-x)=f(x)g(x)=(fg)(x),$ which implies that $fg$ is even. Similarly, let $f$ and $g$ be odd functions. Then, $f(-x)=-f(x)$ and $g(-x)=-g(x)$ for all $x$ in the domain of $f$ and $g$. So we have $(fg)(-x)=f(-x)g(-x)=(-f(x))(-g(x))=f(x)g(x)=(fg)(x),$ which implies that $fg$ is odd. Finally, let $f$ be even and $g$ be odd. Then $f(-x)=f(x)$ and $g(-x)=-g(x)$ for all $x$ in the domain of $f$ and $g$. So we have $(fg)(-x)=f(-x)g(-x)=f(x)(-g(x))=-f(x)g(x),$ which is equal to $-(fg)(x)$. Therefore, $fg$ is neither even nor odd. Step by Step Solution: Step 1. Use the definition of an even function to prove that if $f$ and $g$ are even functions, then $fg$ is even. Step 2. Use the definition of an odd function to prove that if $f$ and $g$ are odd functions, then $fg$ is odd. Step 3. Use the definitions of even and odd functions to prove that if $f$ is even and $g$ is odd, then $fg$ is neither even nor odd. Final Answer: If $f$ and $g$ are both even functions, then $fg$ is even. If $f$ and $g$ are both odd functions, then $fg$ is odd. If $f$ is even and $g$ is odd, then $fg$ is neither even nor odd.