This question is "Given that $H(x) = f(x)-g(x) $ and $f(3) = -4$ and $g(3)=2$ and $h(3)=-2$, which functions f and/or g (both or neither) is/are odd?"
I get completely stuck because $H(x)$ is listed with a capital $H$ and the $h(3)$ value is listed with lower case which makes me think they are not the same function at all.
I would have thought that $H(3) = f(3)-g(3) = -4-2=-6$ which does not equal $h(3)$ which is given as $-2$.
I fully understand that if $f(x)$ is even then $f(-3) = f(3)$ and if $f(x)$ is odd then $f(-3) = - f(3)=4$.
Difficult to see where to go from here.
I will explain it as I suppose it is. I can't be sure of what I'm saying. For me there is a little mistake which is $H\left(-3\right)=-2$ and not $H\left(3\right)=-2$ because as I said, there's no point giving you this because you have already $f(3)$ and $g(3)$. Then you have $$ H\left(3\right)=-6 \text{ and }H\left(-3\right)=f\left(-3\right)-g\left(-3\right)=-2$$
$\bullet$ Suppose that $f$ is even then $$ g\left(-3\right)=2+f\left(3\right)=-2=-g\left(3\right) $$ which means both $f$ and $g$ are even and odd respectively. It is the same if i suppose first $g$ is even. But i've not proved they were actually even or odd, if one is of a kind the other will be the other.
$\bullet$ Suppose that $f$ is odd then $$ g\left(-3\right)=2-f\left(3\right)=6 \ne \pm g\left(3\right) $$ With no more hypothesis on $h$, we can't tell nothing. They could be no odd/even functions too.