$(23)$ If the function $f$ has the rule $f(x) = \sqrt{x^2-9}$ and the function $g$ has the rule $g(x) = x+5$.

$a. $ find integers $c$ and $d$ such that $f(g(x)) =\sqrt{(x+c)(x+d)}$

$b.$ state the maximal domain for which $f(g(x)) $ is defined

For question $(23) \;b.$, in the above , I am struggling with finding the domain. The answer states that $x\le8$ and $x\ge2$.

However, for $f(x)$, you can’t have a negative square root over the real field so $x\le -3$ and $x\gt 3$. So because x has already got this restricted domain, I had gotten the answer that $f(g(x))$ would mean $x\le$ and $x\gt3$...

I know this is a very basic question but I just need some clarification.


  • $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures $\endgroup$ Jun 5 '18 at 13:31
  • $\begingroup$ Is it possible to do so on mobile? $\endgroup$ Jun 5 '18 at 13:33
  • $\begingroup$ it indeed is , but itll be a bit hard. I'll edit this one for you but next time remember to use mathjax. Help us help you $\endgroup$ Jun 5 '18 at 13:34

I think you must have made a mistake at ($a$) or haven't noticed how $f$ and $f\circ g$ may have different definition domains. You noticed how $f$ is defined except for $-3 \leq x\leq 3$. Then $f\circ g$ is defined for $-3\leq g(x) \leq 3$ which is $-8\leq x \leq -2$.

You may find that $f(g(x))= \sqrt{(x+2)(x+8)}$.

Then, as you stated we must have $(x+2)(x+8)\geq 0$ for $f\circ g$ to be defined. This means that $(x+2)$ and $(x+8)$ have the same sign which happens for $x\leq -8$ or $x\geq -2$ The maximal definition domain is $(-\infty,-8)\cup(-2,\infty)$.

The $-$ signs may come from an error in your question (for instance is it $g(x)=x+5$ or $g(x)=x-5$?).


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