If $f$ and $g$ are both odd, show that:
a) $f+g$ is odd
b) $f-g$ is odd
c) $fg$ is even
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2How do you plan to handle that ? – Oct 11 '15 at 16:42
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Welcome to the Math Stack Exchange. What have you tried so far? – Paul Sundheim Oct 11 '15 at 16:44
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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.
layman
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@ChrisSmith I should have mentioned a function is even if $h(-x) = h(x)$ for every real number/input $x$. So for part $c$, you need to show if $f$ and $g$ are both odd, then $fg$ is even, i.e., $(fg)(-x) = (fg)(x)$ for every real number $x$ (and this end result is different than what you need to show for parts a and b). – layman Oct 11 '15 at 17:27
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For arbitrary constants $a,b$
$af(-x)+bg(-x)=-\{af(x)+bg(x)\}$
$f(-x)\cdot g(-x)=-f(x)\{-g(x)\}$
lab bhattacharjee
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