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!
Best Answer
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 $[0,\sqrt 3).$