Intervals and Solving Inequalities

An interval is a special kind of subset of the real numbers R, consisting of all real numbers lying between two fixed bounds. Six standard types:

1. OPEN INTERVAL: for $a,b\in R$ with $a<b$, the set $\{x/x\in R, a<x<b\}$ is the open interval $(a,b)$ — all numbers strictly between a and b belong to it, but a and b themselves do NOT (Fig. 5.8).
2. CLOSED INTERVAL: the set $\{x/x\in R, a\le x\le b\}$ is the closed interval $[a,b]$ — this time a and b themselves ARE included, along with everything between them (Fig. 5.9).
3. SEMI-CLOSED INTERVAL: $[a,b) = \{x/x\in R, a\le x<b\}$ — a is included, but b is not (Fig. 5.10).
4. SEMI-OPEN INTERVAL: $(a,b] = \{x/x\in R, a<x\le b\}$ — a is excluded, but b is included (Fig. 5.11).
5. The set of all reals greater than a is the ray $(a,\infty)=\{x/x\in R, x>a\}$ (Fig. 5.12); the set of all reals greater than or equal to a is $[a,\infty)=\{x/x\in R, x\ge a\}$ (Fig. 5.13).
6. Similarly, the set of all reals less than b is $(-\infty,b)=\{x/x\in R, x<b\}$ (Fig. 5.14); less than or equal to b is $(-\infty,b]=\{x/x\in R, x\le b\}$ (Fig. 5.15).
7. The entire set of real numbers is $R=(-\infty,\infty)=\{x/x\in R, -\infty<x<\infty\}$ (Fig. 5.16). Note $\infty$ and $-\infty$ are never themselves included, so the bracket beside them is always round.

RULES OF INEQUALITY used to solve inequalities: (1) if $0<a<b$ and $k>0$, then $a\pm k<b\pm k$, $ka<kb$, and $\frac{a}{k}<\frac{b}{k}$ — adding, subtracting, or multiplying/dividing by a POSITIVE number preserves the inequality direction. (2) if $0<a<b$ and $k<0$, then $ka>kb$ and $\frac{a}{k}>\frac{b}{k}$ — multiplying or dividing by a NEGATIVE number REVERSES the inequality direction.

SOLVING A LINEAR COMPOUND INEQUALITY, e.g. $-7<2x+5\le9$: subtract 5 throughout to get $-12<2x\le4$, then divide by 2 to get $-6<x\le2$, so $x\in(-6,2]$.

SOLVING A QUADRATIC INEQUALITY by sign analysis, e.g. $x^2+2x<15$: rewrite as $x^2+2x-15<0$, factor as $p(x)=(x-3)(x+5)<0$. The critical points $x=3,-5$ (where the factors are zero) divide the number line into three regions: $x<-5$, $-5<x<3$, $x>3$. Testing one representative point from each region (e.g. $x=-6, 0, 4$) and checking the sign of $p(x)$ shows $p(x)>0$ for $x<-5$, $p(x)<0$ for $-5<x<3$, and $p(x)>0$ for $x>3$. Since we need $p(x)<0$, the solution is $-5<x<3$, i.e. $x\in(-5,3)$. …