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I am in need of some help. Right now I am doing some test corrections for my Honors High School Pre-calc class. This class is Algebra-based calculus. So the question is:

Find $(f\circ g)(x)$ and its domain when $f(x)=x^2+9$ and $g(x)=\sqrt{x+3}$.

I solved $(f\circ g)(x)$ to get $x+12$. So I understand how to get $(f\circ g)(x)$, but I do not understand how to get the domain. According to my test, it says that answer for the domain is $[-3,\infty)$. When I graph it on Desmos.com and my TI-84 Plus I get a straight line, so I really don't know how they got that answer. I would really appreciate it if someone could help me! Thanks in advance!

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The domain of this function is indeed $[-3, \infty)$, assuming $\sqrt[]{x}$ is undefined when $x<0$.

This is because $$(f\circ g)(x) = \left[\sqrt[]{x+3}\right]^2+3.$$ Notice that when $x < -3$, we are taking the square root of a negative number, which is undefined. Therefore, the only values of $x$ for which $(f\circ g)(x)$ is defined are those satisfying $x \geq -3$.

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  • $\begingroup$ Thank you very much, @Austin Weaver! That makes sense now. $\endgroup$ – BlazeRyder Oct 2 '17 at 21:11
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Let $D(f)$ be a domain of the function $f$.

Thus, by definition we have: $$D(f\circ g)=D(g)\cap\{x|g(x)\in D(f)\}$$ In our case $$f(g(x))=\left(\sqrt{x+3}\right)^2+9,$$ $$D(f)=\mathbb R$$ and $$D(g)=[-3,+\infty).$$

Thus, the domain is $[-3,+\infty)$.

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    $\begingroup$ +1 from me as compensation. I don't know who and why downvoted your answer as it is exactly same as accepted answer and came even earlier $\endgroup$ – Vidyanshu Mishra Oct 2 '17 at 21:11
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    $\begingroup$ @VidyanshuMishra The accepted answer has explanation and pedagogy, rather than just showing off that the answerer knows the answer. Hence my downvote. $\endgroup$ – T. Bongers Oct 2 '17 at 21:34
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    $\begingroup$ @MichaelRozenberg An explanation would be something like Austin Weaver's answer, that talks about why the domain is what it is and how you found it, rather than just some notation (that the asker probably has no idea what it is) and "Thus the answer is..." Compare your answer with the answer that actually talks about what the issues are. $\endgroup$ – T. Bongers Oct 3 '17 at 2:15
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    $\begingroup$ @MichaelRozenberg I'm not the asker, and I do understand your answer. I just don't think it's a very good one, because it doesn't have even a single written sentence explaining why it's the correct answer. Answers should be reasonably self-contained, and not need follow-up questions from the poster. $\endgroup$ – T. Bongers Oct 3 '17 at 14:39
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    $\begingroup$ I already explained why: I downvoted your answer because I do not believe it is a high-quality answer that teaches something. It just states an answer and almost nothing else. This is also far from the first time that I've left comments to this effect on your posts and had this conversation, so next time I'll just vote and move on. (And "doing test corrections" almost surely means that the asker is a student correcting their own work, not a teacher grading.) $\endgroup$ – T. Bongers Oct 3 '17 at 14:54

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