## Thomas' Calculus 13th Edition

$a.\quad$Not necessarily true. $b.\quad$Necessarily true. $c.\quad$Necessarily true. $d.\quad$Necessarily true. $e.\quad$Necessarily true. $f.\quad$Necessarily true. $g.\quad$Necessarily true. $h.\quad$Necessarily true.
In interval notation, the set $\{x|2\lt x \lt 6\}$represents the open interval $I=(2,6), \quad$ containing numbers between 2 and 6 (but without 2 and 6). $a.\quad$ This interval, $(0,4)$ contains numbers between $0$ and $4$ (no borders). There are numbers $x$ from interval $I$, such as $x=5$, for which this statement is not true. And, there are numbers from $I$, such as $x=3$, for which this statement holds true. Thus, it is not necessarily true. $b.\quad$ By rule 1 from the "Rules for Inequalities" table, we can add 2 to each part of the compound inequality: $0+2 \lt x-2+2 \lt 4+2$ $2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I.$ Necessarily true. $c.\quad$ By rule 3 from the "Rules for Inequalities" table, $1 \displaystyle \lt \frac{x}{2} \lt 3,\qquad$ multiply all parts with $2$: $2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I.$ Necessarily true. $d.\quad$ By rule 6 from the "Rules for Inequalities" table, $6\gt x \gt 2$ which is another way of writing $2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I.$ Necessarily true. $e.\quad$ If $x$is taken from the inteerval $I$, it is positive, so we can apply rule 6 from the "Rules for Inequalities" table, $1\displaystyle \gt \frac{x}{6} \gt \frac{1}{3}\quad$ apply rule 3 (multiply with 6) $6\gt x \gt 2$ which is another way of writing $2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I.$ Necessarily true. $f.\quad$ Applying property 5 from the table "Absolute values and intervals", $-2\lt x-4 \lt 2 \quad$ apply rule 1 ( add 4 to each part) $2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I.$ Necessarily true. $g.\quad$ By rule 4, multiplying with $(-1)$ we obtain $6\gt x \gt -2$ which is another way of writing $-2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I$, because $-2\lt 2 \lt x \lt 6$ (The interval $I$ is a subset of this interval) Necessarily true. $h.\quad$ By rule 4, multiplying with $(-1)$ we obtain $6\gt x \gt 2$ which is another way of writing $2 \lt x \lt 6$, which is true for any $x$ we choose from the interval $I.$ Necessarily true.