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So I have two functions. $f(x) = e^{-x^2+1}$ and $g(x)=\sqrt{x^2-4x+3}$. I am then asked to determine the domain and range of

$a)f∘g,$

$b)g∘f$

I already did part $a)$ and the domain for part $b)$.

For part $a)$, the domain was $(-\infty,1)\cup(3,\infty)$ and the range was $(0,e^2)$.

For part $b$, I figured out that the domain was $(-\infty,-1]\cup[1,\infty)$. I am not sure how to find the range though. Normally, I would take the inverse of g∘f and find the domain of that, and although I can do it, I don't think I did it correctly.

Currently, I did figure out that $g∘f$ is $\sqrt{e^{-2x^2+2}-4e^{-x^2+1}+3}$. How do I find the range of this mess though? I attempted to take the inverse which I believe is:

$y=\pm\sqrt{1-\ln(2\pm\sqrt{1+y^2})}$

Although I know that Wolfram Alpha is not the arbitrator of what correct is, it's generally been right and my answer disagrees with what Wolfram alpha has obtained (As seen here). In addition, the range is something that I am not sure how Wolfram obtained (As seen here). This also looks REALLY messy.

Can anyone guide me as to how this was obtained? That would be much appreciated!

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  • $\begingroup$ I see how you got $(-\infty,1)\cup(3,\infty)$ (except maybe the choice of open/closed) but not how you got $e^2.$ When you were computing the range, did you forget the gap in the domain? I think that part is easier to solve if you do not work out the formula for $f\circ g.$ $\endgroup$ – David K Sep 11 at 11:58
  • $\begingroup$ I took the inverse of the function and found the domain of that to get the range. $\endgroup$ – Future Math person Sep 11 at 18:23
  • $\begingroup$ That seems like an unnecessarily difficult way to do it. Also error-prone, since (I think) it gives the wrong answer. What value of $x$ satisfies $f(g(x))=e^{3/2},$ for example? $\endgroup$ – David K Sep 11 at 18:42
  • $\begingroup$ Really? I didn't think it was too bad. I ended up getting $f(g(x))=-2 \pm \sqrt{2-\ln(x)}$. This would mean $2-\ln(x) \geq 0$ and thus $e^2 \geq x$ $\endgroup$ – Future Math person Sep 11 at 18:44
  • $\begingroup$ Also I get $g(1)=g(3)=0,$ and $0$ is in the domain of $f,$ so I would not exclude $1$ and $3$ from the domain in part a). I think you may need to be a lot more careful about checking what happens at your boundaries. (This is relevant to the range as well.) $\endgroup$ – David K Sep 11 at 18:47
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To find the range you want, namely of $g(f(x))=F(x),$ note that $F(x)$ is never negative. Also, it is defined and continuous at all real $x.$ Then it is an even function of $x.$ Thus, it suffices to consider only the range for positive $x,$ say.

Note that $F(0)>0.$ Also, as $x\to +\infty,$ we have $F(x)\to \sqrt 3.$ Thus, the range is at least $[F(0),\sqrt 3),$ by IVT. It only remains to see if the function attains $\sqrt 3$ at any point, or if it ever falls below $F(0).$ The first is easily answered in the negative (set $F(x)=\sqrt 3$ to deduce a contradiction), so the range is half-open. So does the function ever go below $e^2-4e+3$? In particular we may ask whether the function ever attains the minimum possible here, namely $0.$ Thus, setting $F(x)=0,$ we see that we want to see if there are real solutions to the quadratic equation $p^2-4p+3=0,$ where $p=e^{-x^2+1}.$ This has solutions. In particular, we get that $e^{-x^2+1}=1,$ or in other words that $$-x^2+1=0.$$ Thus $F(x)$ vanishes in at least two points.

It therefore is the case that the range is $[0,\sqrt 3).$

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Note that the range of $f$ on $(-\infty,-1]\cup[1,\infty)$ is $(0,1]$. The range of $g$ on $(0,1]$ is $[0,\sqrt{3})$.

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Let $f,g$ be given by $$ \begin{cases} f(x)=e^{1-x^2}\\[4pt] g(x)=\sqrt{x^2-4x+3}\\ \end{cases} $$ $\bullet\;\,$Part $(\mathbf{a}){\,:}\;\,$Find the domain and range of $f\circ g$.

    Since the domain of $f$ is $\mathbb{R}$, the domain of $f\circ g$ is the same as the domain of $g$.

    Hence the domain of $f\circ g$ is $(-\infty,1]\cup [3,\infty)$.

    Since $f$ is an even function, the range of $f$ is the same as the range of $f$ on the restricted domain $[0,\infty)$.

    On the interval $[0,\infty)$, $f$ is strictly decreasing, and $f$ approaches zero from above as $x$ approaches infinity.

    Since $f$ is continuous, it follows that the range of $f$ is $(0,e]$.

    For $x\ge 3$, $g$ realizes all values in $[0,\infty)$, so the range of $f\circ g$ is the same as the range of $f$.

    Hence the range of $f\circ g$ is $(0,e]$.

$\bullet\;\,$Part $(\mathbf{b}){\,:}\;\,$Find the domain and range of $g\circ f$.

    The domain of $g\circ f$ is the set of all real $x$ such that $f(x)$ is in the domain of $g$, which is the set of all real $x$ such that $f(x)\le 1$ or $f(x)\ge 3$.

    But we can't have $f(x)\ge 3$, since the maximum value of $f$ is $e$.

    For the condition $f(x)\le 1$, we get $$f(x)\le1\iff e^{1-x^2}\le 1\iff 1-x^2\le 0\iff |x| \ge 1$$ hence the domain of $g\circ f$ is $(-\infty,-1]\cup [1,\infty)$.

    Restricted to the domain $(-\infty,-1]\cup [1,\infty)$, the range of $f$ is $(0,1]$, hence the range of $g\circ f$ is the range of $g$ on the restricted domain $(0,1]$.

    It follows that the range of $g\circ f$ is $[0,\sqrt{3})$.

$\bullet\;\,$To summarize the results:

  • The domain of $f\circ g$ is:$\;\,(-\infty,1]\cup [3,\infty)$.
  • The range of $f\circ g$ is:$\;\,(0,e]$.$\\[6pt]$
  • The domain of $g\circ f$ is:$\;\,(-\infty,-1]\cup [1,\infty)$.
  • The range of $g\circ f$ is:$\;\,[0,\sqrt{3})$.
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