evaluating positive and negative regions for a simple compound function

Question

I'm trying to understand how to analytically determine the regions where $y_1 \lt y_2$. If I set the functions equal to each other it yields $x=14$ which is true for that point but, as you can see from the graph, that doesn't help define a region. The region (from the graph) is between $-1$ and $4$ which are the two vertical asymptotes but that is by graph inspection rather than analysis. Thanks. Chris

Answer 1

As regions you can use the open intervals where both functions are defined and are not equal.

That is, find all the points where

• either $y_1$ or $y_2$ are not defined, or
• $y_1$ and $y_2$ are equal

and remove them from the real numbers. In each one of the intervals that you get, $y_1$ and $y_2$ are continuous. So is their difference $y_3 = y_1 - y_2$. And since $y_3$ does not vanish, it follows that $y_3$ has a constant sign at each of those intervals. To find the sign at an interval, all you have to do is look at the sign of $y_3(x)$ for some $x$ in that interval.

Answer 2

there are $4$ potentially relevant points - poles $x=-1, x=4$, intersections $x=14$ and infinity. Of each of these points you have to check slightly to the left and to the right the inequality and be sure it holds everywhere in between. In particular there is the region $(14,\infty)$ which you probably overlooked.