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

Key Concepts: Even Function, Odd Function, Function Addition, Justification Explanation: An even function $f$ satisfies $f(-x)=f(x)$ for all $x$, while an odd function $g$ satisfies $g(-x)=-g(x)$ for all $x$. To determine whether $f+g$ is even or odd, we evaluate $(f+g)(-x)$ and $(f+g)(x)$ and compare the results. Step by Step Solution: Step 1. Assume $f$ and $g$ are even functions. Step 2. Then $f(-x)=f(x)$ and $g(-x)=g(x)$ for all $x$. Step 3. Therefore, $(f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x)$ for all $x$, so $f+g$ is even. Step 4. Assume $f$ and $g$ are odd functions. Step 5. Then $f(-x)=-f(x)$ and $g(-x)=-g(x)$ for all $x$. Step 6. Therefore, $(f+g)(-x)=f(-x)+g(-x)=-f(x)-g(x)=-(f+g)(x)$ for all $x$, so $f+g$ is odd. Step 7. Assume $f$ is even and $g$ is odd. Step 8. Then $f(-x)=f(x)$ and $g(-x)=-g(x)$ for all $x$. Step 9. Therefore, $(f+g)(-x)=f(-x)+g(-x)=f(x)-g(x)=(f+g)(x)$ for all $x$, so $f+g$ is even. Step 10. Therefore, $f+g$ is even when $f$ and $g$ are both even, even when $f$ is even and $g$ is odd, and odd when $f$ and $g$ are both odd. Step 11. Justification: The proof in steps 3 and 6 uses the properties of even and odd functions as defined in the problem. Final Answer: If $f$ and $g$ are both even functions, then $f+g$ is even. If $f$ and $g$ are both odd functions, then $f+g$ is odd. If $f$ is even and $g$ is odd, then $f+g$ is even.