This is a question by PapaSmurf:

Quote:

2)

Letandf(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]

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- May 14th 2011, 10:44 AMtopsquarkComposition of Two Functions
This is a question by PapaSmurf:

Quote:

__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]

- May 14th 2011, 10:55 AMTheEmptySet
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. - May 14th 2011, 11:06 AMPapaSmurf
Awesome, thank you :)

But, I don't completely understand what you mean at the end.

Quote:

Now just solve for k and multiply the two solutions together.

but don't understand where to go from there... would that be my answer? - May 14th 2011, 11:26 AMTheEmptySet
In the original problem statement is says find the product all of distinct k.

Well you have the two distinct values of k so find their product so multiply them together.