# Domain of composite function with initially restricted domain

$$(23)$$ 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.

Thanks!

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures Jun 5 '18 at 13:31
• Is it possible to do so on mobile? Jun 5 '18 at 13:33
• 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 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 $g(x)=x-5$?).