Find upper and lower bound 
$\sqrt 3 = 1.73$ is correct to 3 significant figures. Find the upper and lower bounds of  $$\frac{\sqrt 3-1}{2+\sqrt 3}$$ rounding your answers to 3 significant figures.


I'm confused here as:
UB of √3=1.735
LB of √3=1.725
I think:
UB of (√3-1)/(2+√3)  =(1.735-1)/(2+1.725)=0.1973…=0.197 (3sf)
LB of (√3-1)/(2+√3)  =(1.725-1)/(2+1.735)=0.1941…=0.194 (3sf)
What I'm confused with is if e.g. to find UB of fraction, I want to do maximum/minimum.
But surely, once I've said I've taken the maximum 1.735, and if that is indeed the value of √3, then I should also use that 1.735 in denominator (and not 1.725). It doesn't make sense to me in this context why we should take 1.725 in the denominator.
As there is no way the value given the approximation to √3 could ever be 0.1973....so it seems silly to have (1.735-1)/(2+1.725).....not saying though it should be (1.735-1)/(2+1.735) (it could be some other value in the range 1.725 to 1.735, but that the value should be consistent??
Or are we in such questions, just taking the worst case and best case scenarios, even if they are in no way plausible?
Not sure if my question makes sense...
 A: So you don't know much about $\alpha=\sqrt{3}$ except that $1.725\le\alpha\le 1.735$, and you are asked to estimate $\beta=\frac{\alpha-1}{2+\alpha}$. In other words, to produce some hard bounds $L$ and $U$ such that you can prove $L\le\beta\le U$, and hopefully $U-L$ is not "too big".
Surely: $\frac{1.725-1}{2+1.735}\le\frac{\alpha-1}{2+\alpha}\le\frac{1.735-1}{2+1.725}$, i.e. $0.1941\ldots\le\beta\le 0.1973\ldots$ so you have got some upper and lower bounds.
You can also break down $\beta=\frac{\alpha-1}{2+\alpha}=\frac{\alpha+2-3}{2+\alpha}=1-\frac{3}{2+\alpha}$. In that case: $1-\frac{3}{2+1.725}\le\beta\le 1-\frac{3}{2+1.735}$, i.e. $0.19463\ldots\le\beta\le 0.19678\ldots$. You have got different (and actually tighter) upper and lower bounds.
You can also get some estimates if you rationalize the expression for $\beta$: $\beta=\frac{\alpha-1}{2+\alpha}=\frac{(\alpha-1)(2-\alpha)}{(2+\alpha)(2-\alpha)}=3\alpha-5$ (using $\alpha^2=3$), so $3\cdot 1.725-5\le\beta\le 3\cdot 1.735-5$, i.e. $0.175\le\beta\le 0.205$ - now those are very much looser than the first ones.
All of those solutions do the job in some sense, because, of course, $\beta=0.19615\ldots$ and it fits all of those intervals. Now, which one would be admitted in your class/your exam etc.? I don't know and I cannot advise you about that. Looks like the first approach is a standard procedure that is expected to be known by the students. The second and third procedure rely on transforming the expression for $\beta$, which may not be possible in general case. Also not all transforms are born equal: the second procedure yielded better estimate than the first, so I believe it would be unfair to be marked down. I am not sure, if someone came up with the third solution, how this should be marked: the estimate is way worse (by an order of magnitude) than even the "standard" first solution.
Outside of school, in real life, of course, you will be looking for the best estimate you can find, with the information given, within the time you've got allocated. You will be judged based on whether what you came up with is good enough for some purpose. Depending on the circumstances, even a fast-and-loose estimate ("worst case" or whatever you may want to call it) may go a long way.
