Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\le-\dfrac{23}{9} \text{ or } x\ge3$
$\bf{\text{Solution Outline:}}$ To find all $x$ such that $f(x) \ge25 ,$ with $f(x)= |2-9x| ,$ solve the inequality \begin{array}{l}\require{cancel} |2-9x|\ge25 .\end{array} Then graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 2-9x\ge25 \\\\\text{OR}\\\\ 2-9x\le-25 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 2-9x\ge25 \\\\ -9x\ge25-2 \\\\ -9x\ge23 \\\\\text{OR}\\\\ 2-9x\le-25 \\\\ -9x\le-25-2 \\\\ -9x\le-27 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to \begin{array}{l}\require{cancel} -9x\ge23 \\\\ x\le\dfrac{23}{-9} \\\\ x\le-\dfrac{23}{9} \\\\\text{OR}\\\\ -9x\le-27 \\\\ x\ge\dfrac{-27}{-9} \\\\ x\ge3 .\end{array} Hence, the solution set is $x\le-\dfrac{23}{9} \text{ or } x\ge3 .$