Odd and even functions, sum, difference and product

Question

If $f$ and $g$ are both odd, show that:

a) $f+g$ is odd

b) $f-g$ is odd

c) $fg$ is even

Answer 1

For arbitrary constants $a,b$

$af(-x)+bg(-x)=-\{af(x)+bg(x)\}$

$f(-x)\cdot g(-x)=-f(x)\{-g(x)\}$

Answer 2

What does it mean for $f(x)$ and $g(x)$ to be odd?

It means $f(-x) = -f(x)$ and $g(-x) = -g(x)$ for every real number/input $x$.

Now, to show, for example, $f + g$ is odd, you need to show for every real number $x$, $(f + g)(-x) = -(f+ g)(x)$.

I'll help you with this one, and the others are left to you:

$(f + g)(-x) = f(-x) + g(-x)$ by the definition of $f + g$, right? But $f$ is odd, and so is $g$, so the above equals $(-f(x)) + (-g(x))$. Then we can factor out the $-$ sign to get that it equals $-(f(x) + g(x))$.

So, we got $(f + g)(-x) = -((f+g)(x))$, which means $f + g$ is odd.