Answer
$\text{a) }
f(g(2))=\dfrac{19}{10}
\\\\\text{b) }
g(f(2))=\dfrac{9}{5}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
With
\begin{array}{l}\require{cancel}
f(x)=
\dfrac{1}{2}x+\dfrac{1}{5}
\\g(x)=
2x-\dfrac{3}{5}
,\end{array}
to find $
f(g(2))
,$ find first $
g(2)
.$ Then substitute the result in $f.$
To find $
g(f(2))
,$ find first $
f(2)
.$ Then substitute the result in $g.$
$\bf{\text{Solution Details:}}$
a) Replacing $x$ with $
2
$ in $g$ results to
\begin{array}{l}\require{cancel}
g(x)=2x-\dfrac{3}{5}
\\\\
g(2)=2(2)-\dfrac{3}{5}
\\\\
g(2)=4-\dfrac{3}{5}
\\\\
g(2)=\dfrac{20}{5}-\dfrac{3}{5}
\\\\
g(2)=\dfrac{17}{5}
.\end{array}
Replacing $x$ with the result above in $f$ results to
\begin{array}{l}\require{cancel}
f(x)=\dfrac{1}{2}x+\dfrac{1}{5}
\\\\
f\left( \dfrac{17}{5} \right)=\dfrac{1}{2}\left( \dfrac{17}{5} \right)+\dfrac{1}{5}
\\\\
f\left( \dfrac{17}{5} \right)=\dfrac{17}{10}+\dfrac{1}{5}
\\\\
f\left( \dfrac{17}{5} \right)=\dfrac{17}{10}+\dfrac{2}{10}
\\\\
f\left( \dfrac{17}{5} \right)=\dfrac{19}{10}
.\end{array}
Hence, $
f(g(2))=\dfrac{19}{10}
.$
b) Replacing $x$ with $
2
$ in $f$ results to
\begin{array}{l}\require{cancel}
f(x)=\dfrac{1}{2}x+\dfrac{1}{5}
\\\\
f(2)=\dfrac{1}{2}(2)+\dfrac{1}{5}
\\\\
f(2)=1+\dfrac{1}{5}
\\\\
f(2)=\dfrac{5}{5}+\dfrac{1}{5}
\\\\
f(2)=\dfrac{6}{5}
.\end{array}
Replacing $x$ with the result above in $g$ results to
\begin{array}{l}\require{cancel}
g(x)=2x-\dfrac{3}{5}
\\\\
g\left( \dfrac{6}{5} \right)=2\left( \dfrac{6}{5} \right)-\dfrac{3}{5}
\\\\
g\left( \dfrac{6}{5} \right)=\dfrac{12}{5}-\dfrac{3}{5}
\\\\
g\left( \dfrac{6}{5} \right)=\dfrac{9}{5}
.\end{array}
Hence, $
g(f(2))=\dfrac{9}{5}
.$
Therefore,
\begin{array}{l}\require{cancel}
\text{a) }
f(g(2))=\dfrac{19}{10}
\\\\\text{b) }
g(f(2))=\dfrac{9}{5}
.\end{array}
