Answer
$x=\left\{ -\dfrac{7}{3},-\dfrac{1}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Given that $
f(x)=\left|\dfrac{3x+4}{3} \right|
,$ to find $x$ for which $
f(x)=1
,$ use substitution. Then use the definition of absolute value equality. Finally, use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Replacing $f(x)$ with $
1
,$ then
\begin{array}{l}\require{cancel}
f(x)=\left|\dfrac{3x+4}{3} \right|
\\\\
1=\left|\dfrac{3x+4}{3} \right|
\\\\
\left|\dfrac{3x+4}{3} \right|=1
.\end{array}
Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{3x+4}{3}=1
\\\\\text{OR}\\\\
\dfrac{3x+4}{3}=-1
.\end{array}
Solving each equation above results to
\begin{array}{l}\require{cancel}
3\cdot\dfrac{3x+4}{3}=3\cdot1
\\\\
3x+4=3
\\\\
3x=3-4
\\\\
3x=-1
\\\\
x=-\dfrac{1}{3}
\\\\\text{OR}\\\\
\dfrac{3x+4}{3}=-1
\\\\
3\cdot\dfrac{3x+4}{3}=3\cdot(-1)
\\\\
3x+4=-3
\\\\
3x=-3-4
\\\\
3x=-7
\\\\
x=-\dfrac{7}{3}
.\end{array}
Hence, $
x=\left\{ -\dfrac{7}{3},-\dfrac{1}{3} \right\}
.$