Find the product of $f(x)$ and $g(x)$ given one of them I'm given the final answer which is $$(g \cdot f)(x) = \frac{1}{x^2+4}\;.$$
Also, i'm given $f(x) = x^2+1$.
I've solved this using the composition, however the second part of the question asks me to find the $g(x)$ which would make this multiplication true. How would I do this? Do I divide the final answer by $f(x)$? 
 A: Yes, becasue by definition $(g\cdot f)(x)=g(x)\cdot f(x)$.
Hence $g(x)=\frac1{(x^4+1)(x^2+1)}$. 
If we had $(g\circ f)(x)=\frac1{x^4+1}$ instead, one possible $g$ would be $g(x)=\frac1{(x-1)^2+4}$.
A: So you are told that $(g \times f)(x) = \dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.
$(g \times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \times f)(x)$ really means $g(x)f(x)$.
So we know what this product is.  It is $g(x)f(x) = \dfrac{1}{x^{2} + 4}$.  We also know that $f(x) = x^{2} + 1$.  So that means:
$$g(x)(x^{2} + 1) = \dfrac{1}{x^{2} + 4} $$
and to solve for $g(x)$, just divide both sides by $x^{2} + 1$ to get:
$$g(x) = \dfrac{\left (\frac{1}{x^{2} + 4} \right )}{x^{2} + 1} $$
Now, how do we simplify this?  Well, $x^{2} + 1$ is the same as $\dfrac{x^{2} + 1}{1}$, so the fraction is really $$\dfrac{\left (\frac{1}{x^{2} + 4} \right )}{\left (\frac{x^{2} + 1}{1}\right)} $$
and when we divide two fractions, we invert the bottom one and multiply, so we get:
$$\dfrac{1}{x^{2} + 4} \cdot \dfrac{1}{x^{2} + 1} $$
And this is just $\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is your final answer (unless you want to multiply the denominator out using the FOIL method).
