Solving Rational Inequalities Analytically

Exercise 5.

Find the solutions of the inequality

$\begin{displaymath}\frac{x^2-9}{x+3}\leq 0.\end{displaymath}$

Answer.

The numerator is zero for $x=\pm 3$ , while the denominator is zero at x=-3; thus we have two critical points:

The tricky part is reading off all solutions. We have to exclude x=-3, since the expression on the left side of the inequality is undefined at x=-3, but points in the "neighborhood" of x=-3 are okay. So the set of solutions is the set $(-\infty,-3)\cup (-3,3]$ .

Note that

$\begin{displaymath}\frac{x^2-9}{x+3}=x-3,\end{displaymath}$

only when $x\not=-3$ . When x=-3, the left side is undefined, while the right side evaluates to -6.

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1998-06-16