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EDIT: turns out $\sqrt{\sqrt{x^{2}+3}-3}$ was actually right. In that case, that gives me a follow-up question of sorts.

So the larger context of the original question was to find the domain of $(f \circ g)(x)$. Following the definition of the domain of $(f \circ g)(x)$ in my textbook (all $x$ in the domain of $g(x)$ such that $g(x)$ falls into the domain of $f(x)$), I first found the domain and range of $f(x)=\sqrt{x-3}$, which was $[3,+\infty)$ and $[0,+\infty)$ respectively. I next found the domain and range of $g(x)=\sqrt{x^{2}+3}$, which was $\mathbb{R}$ and $[\sqrt{3},\infty)$ respectively.

From there, I reasoned that since the range of $f(x)$ falls within the domain of $g(x)$, the domain of $f(x)$, $[3,+\infty)$ must also be the domain of $(f \circ g)(x)$. A graph of $\sqrt{\sqrt{x^{2}+3}-3}$ indicates otherwise. Thus, my (real, presumably) question is why $[3,+\infty)$ was the incorrect domain of $(f \circ g)(x)$.

ORIGINAL QUESTION

I was asked to evaluate $(f \circ g)(x)$ given the following functions: $$f(x)=\sqrt{x-3}$$ $$g(x)=\sqrt{x^{2}+3}$$ My first answer was that $(f \circ g)(x)=\sqrt{\sqrt{x^{2}+3}-3}$, which would make sense considering $(f \circ g)(x)=f(g(x))$. However, this is apparently not correct (graphing $f(g(x))$ as $Y_{1}(Y_{2}(X))$ in my calculator didn't match graphing $\sqrt{\sqrt{x^{2}+3}-3}$). Why?

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    $\begingroup$ Maybe you have a Polish calculator $\endgroup$ – Paolo Leonetti Jun 30 '17 at 22:28
  • $\begingroup$ You're right and the machine is wrong. Now celebrate ! $\endgroup$ – krirkrirk Jun 30 '17 at 22:30
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To define $f (g (x)) $, we need that

$$g (x)\ge 3$$ or $$\sqrt {x^2+3}\ge 3$$

which is equivalent to $x^2\ge 6$

or $$x\in (-\infty,-\sqrt {6}]\cup [\sqrt {6},+\infty) .$$

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This is exactly correct (for numbers for which the roots are defined). Maybe double-check your calculator work.

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