Answer
$f\left( g\left( x \right) \right)$ and $g\left( f\left( x \right) \right)$ of the function $f\left( x \right)=\frac{3}{5}x+\frac{1}{2}$ and $g\left( x \right)=\frac{5}{3}x-2$ is $f\left( g\left( x \right) \right)=x-\frac{7}{10}\text{ and }g\left( f\left( x \right) \right)=x-\frac{7}{6}$ respectively, and f and g are not inverses of each other.
Work Step by Step
Now, consider the function:
$\begin{align}
& f\left( g\left( x \right) \right)=f\left( \frac{5}{3}x-2 \right) \\
& =\frac{3}{5}\left( \frac{5}{3}x-2 \right)+\frac{1}{2} \\
& =\frac{15}{15}x-\frac{6}{5}+\frac{1}{2} \\
& =x-\frac{12-5}{10}
\end{align}$
That is,
$\begin{align}
& f\left( g\left( x \right) \right)=x-\frac{12-5}{10} \\
& =x-\frac{7}{10}
\end{align}$
Next considering
$\begin{align}
& g\left( f\left( x \right) \right)=g\left( \frac{3}{5}x+\frac{1}{2} \right) \\
& =\frac{5}{3}\left( \frac{3}{5}x+\frac{1}{2} \right)-2 \\
& =x+\frac{5}{6}-2 \\
& =x-\frac{7}{6}
\end{align}$
This shows that the functions $f$ and $g$ are not inverses of each other, since none of the compositions resulted in x.
Hence, $f\left( g\left( x \right) \right)$ and $g\left( f\left( x \right) \right)$ of the function $f\left( x \right)=\frac{3}{5}x+\frac{1}{2}$ and $g\left( x \right)=\frac{5}{3}x-2$ is $f\left( g\left( x \right) \right)=x-\frac{7}{10}\text{ and }g\left( f\left( x \right) \right)=x-\frac{7}{6}$ respectively, and f and g are not inverses of each other.
