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 $f(x)$?

  • $\begingroup$ is that $$(f\cdot g)(x)=\frac1{x^2+4}$$ or $$(f\cdot g)(x)=\frac1{x^2}+4\;?$$ Either way, the answer to your question is yes. $\endgroup$ – Brian M. Scott Apr 19 '16 at 21:34
  • $\begingroup$ the first one, and how would i go about doing that? I know once I divide i take the reciprocal of g(x) and multiply using the FOIL method, however my answer makes no sense $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:35
  • $\begingroup$ thank you for the edit, the question is now formatted correctly. My math is however not giving me the correct answer, could someone guide me through it? $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:35
  • $\begingroup$ Your best bet is to learn to use basic MathJax; there’s a tutorial here. Until then, be sure to use enough parentheses to make your expressions completely unambiguous. $\endgroup$ – Brian M. Scott Apr 19 '16 at 21:37
  • $\begingroup$ For the division, just do it: $$\frac{\frac1{x^2+4}}{x^2+1}=\frac1{x^2+4}\cdot\frac1{x^2+1}=\ldots$$ $\endgroup$ – Brian M. Scott Apr 19 '16 at 21:38

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 $g(x)=\frac1{(x-1)^2+4}$.

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  • $\begingroup$ when taking the reciprocal and multiplying, I get a quartic function which makes no sense $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:39
  • $\begingroup$ @SaadSiddiqui $\frac1{(x^4+4)(x^2+1)}\cdot(x^2+1)=\frac1{x^4+4}$ $\endgroup$ – Hagen von Eitzen Apr 19 '16 at 21:40
  • $\begingroup$ Why does a quartic function make no sense?... or really a ratio of 1/ a quartic function. $\endgroup$ – Doug M Apr 19 '16 at 21:40
  • $\begingroup$ well when multiplying out the denominators, I get an answer of x^4 + 5x^2 + 5. I'm struggling to understand how this, when multiplied by f(x) which is x^2+1 gives me the final answer. $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:44

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).

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  • $\begingroup$ I got to this last part, but it's the multiplying out which has stumped me $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:41
  • $\begingroup$ @SaadSiddiqui Ok, so $(x^2 + 4)(x^{2} + 1) = x^{4} + x^{2} + 4x^{2} + 4 = x^{4} + 5x^{2} + 4$, so your final answer should be $\dfrac{1}{x^{4} + 5x^{2} + 4}$. $\endgroup$ – layman Apr 19 '16 at 21:42
  • $\begingroup$ I got the same answer, but how does this (when multiplied with the original f(x) ) give me my final answer? $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:44
  • $\begingroup$ @SaadSiddiqui Well, first you need to understand that $x^{4} + 5x^{2} + 4 = (x^{2} + 4)(x^{2} + 1)$, which we know since we multiplied the right hand side out to get the left hand side. So $\dfrac{1}{x^{4} + 5x^{2} + 4} = \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$. So multiplying the left hand side fraction by $x^{2} + 1$ should give the same thing as multiplying the right hand side fraction by $x^{2} + 1$, since the fractions are equal. But what happens when we multiply the right hand side fraction by $x^{2} + 1$? We get $(x^{2} + 1) \cdot \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is... $\endgroup$ – layman Apr 19 '16 at 21:47
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    $\begingroup$ Yes completely, by not multiplying out the denominator, you can simply cancel out in the next step. Thank you so much! $\endgroup$ – Saad Siddiqui Apr 19 '16 at 21:54

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