This is a question by PapaSmurf:
2)
Let and f(x) = 9x + k and g(x) = kx + 9.
Find the product of all distinct k such that f(g(x)) = g(f(x)). [Word for word, no typos]
Given that
$\displaystyle f(x)=9x+k \quad g(x)=kx+9$
then
$\displaystyle f(g(x))=9(kx+9)+k \quad \text{ and } \quad g(f(x))=k(9x+k)+9$
setting them equal and expanding gives
$\displaystyle 9kx+81+k=9kx+k^2+9 \iff k^2-k-72=0$
Now just solve for k and multiply the two solutions together.