How do I determine the value of x that sets a given range to a function? I had a question on my calculus test which said:
Let $ f(x) $ be a function that is defined by the relation $\frac{3x+2}{x-1}$.
I was able to do all the sub-questions related to it except for one. The question stated that for a number $A$, where $x>A$, $f(x)$ must have the range $]2.9,3.1[$. 
How do I determine the possible values of the real number $A$?
From the graph, I can approximate that when $x>52$, $f(x)$ starts to fall into the given range, but how do I prove this mathematically?
Thank you!
 A: We know that the function values approach $3$ as $x\to \infty$. This question is asking when we get close. Let's solve the equations $f(x)=2.9$ and $f(x)=3.1$
$$\frac{3x+2}{x-1}=2.9\implies x=-49$$
and
$$\frac{3x+2}{x-1}=3.1\implies x=51$$
Outside of those bounds, i.e., when $x<-49$ and when $x>51$, your desired inequality should be satisfied. Because of the way your question is phrased ("find $A$ such that when $x>A$...."), we can just use the $x>51$ condition.

One way to get a better sense of what's going on is to rewrite the function this way (using polynomial long division, if necessary):
$$f(x)=3+\frac{5}{x-1}$$
Looking at it this way, you can see that the question is really asking, when is $\left|\frac{5}{x-1}\right|<\frac{1}{10}$?
A: Note that the function is decreasing (the derivative is $\frac{-5}{(x-1)^2}$) and has a limiting value of $3$ as $x$ approaches $+\infty$. So we can choose $A$ such that $f(A) = 3.1$. This is relatively easy to solve for $A$ -- see G Tony Jacobs's answer, where he does this and gets $A = 51$.
