#### Answer

$-\dfrac{1}{2}\le x \le \dfrac{7}{2}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find all $x$ such that $f(x)
\le4
,$ with $f(x)=
|2x-3|
,$ solve the inequality
\begin{array}{l}\require{cancel}
|2x-3|\le4
.\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 $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-4\le 2x-3 \le4
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-4\le 2x-3 \le4
\\\\
-4+3\le 2x-3+3 \le4+3
\\\\
-1\le 2x \le 7
\\\\
-\dfrac{1}{2}\le \dfrac{2x}{{2}} \le \dfrac{7}{{2}}
\\\\
-\dfrac{1}{2}\le x \le \dfrac{7}{2}
.\end{array}
Hence, the solution set $
-\dfrac{1}{2}\le x \le \dfrac{7}{2}
.$