Solving inequalities comparing $f(x)$ to $0$ where $f$ is an elementary function Any inequality comparing elementary functions can be rearranged to compare some elementary function $f$ to $0$.  What is the best way to approach, in general, solving such inequalities at the precalculus level?
 A: One general technique is called the "test-point method" or the "boundary algorithm."  The idea is that, when we are comparing an elementary function $f$ to $0$, the only places where the comparison can change between being true and false are places where the function $f$ changes sign, and (from the Intermediate Value Theorem), $f$ can only change sign where $f(x)=0$ or where $f$ is discontinuous.  The values of $x$ for which $f(x)=0$ or $f$ is discontinuous (the "boundary" values) partition the real numbers into open intervals where, for each interval, the inequality comparing $f$ to $0$ is either true for the whole interval or false for the whole interval.  So, testing a single value in each interval ("test-point") determines whether that whole interval is or is not part of the solution set.  Whether or not each boundary value is included in the solution set depends on why it is a boundary value and what type of inequality is being solved (it is usually simple to determine which boundary values, if any, are solutions to the inequality by inspection).

As an example, let's solve $\frac{x+4}{x+1}\geq 0$.  The function $f(x)=\frac{x+4}{x+1}$ is zero only when $x=-4$ and discontinuous only when $x=-1$.  Graphing these on a number line gives:
$$\leftarrow\!\!\underset{-4}{-\!\!\bullet\!\!-}\!\!-\!\!\underset{-1}{-\!\!\circ\!\!-}\!\!\rightarrow$$
(open circle at -1 because $x=-1$ makes $f$ undefined, so it cannot be part of the solution set; closed circle at -4 because $f(-4)=0$, which satisfies $f(x)\geq 0$).
Now, there are 3 intervals to test: $(-\infty,-4)$, $(-4,-1)$, and $(-1,\infty)$.  Relatively simple-to-test values in the intervals are -5, -2, and 0, respectively. $\frac{-5+4}{-5+1}=\frac{-}{-}=+\geq 0$; $\frac{-2+4}{-2+1}=\frac{+}{-}=-\not\geq 0$; $\frac{0+4}{0+1}=\frac{+}{+}=+\geq 0$.  So, the solution is $(-\infty,-4]\cup(-1,\infty)$.

It is often the case that, knowing whether a particular interval is or is not part of the solution, oen can determine whether an adjacent interval is or is not part of the solution by looking at what aspect of $f$ caused the boundary point between the two intervals.  That is, it is often possible to tell by inspection whether or not $f$ changes sign across that boundary.  So, testing a point in each interval is not always necessary.
