Do the \(\frac{1}{3}\) and the \(3\) cancel each other? No. The \(3\) is part of the input, affecting \(x\) right away. Then \(g\) does whatever it does to \(3x\text{,}\) and then we multiply the result by \(\frac{1}{3}\text{.}\) Since the function \(g\) acts “in between,” we don’t have the chance to cancel the \(3\) with the \(\frac{1}{3}\text{.}\) Let’s see what actually happens:
Those parentheses encase “\(3x\text{,}\)” so we are meant to treat “\(3x\)” as the input. We will keep the \(\frac{1}{3}\) where it is until it is possible to simplify:
\begin{align*}
\frac{1}{3}g(\phantom{3x})\amp=\frac{1}{3}\left(2(\phantom{3x})^2+8\right)\\
\frac{1}{3}g(\substitute{3x})\amp=\frac{1}{3}\left(2(\substitute{3x})^2+8\right)\\
\amp=\frac{1}{3}\left(2\left(9x^2\right)+8\right)\\
\amp=\frac{1}{3}\left(18x^2+8\right)\\
\amp=6x^2+\frac{8}{3}
\end{align*}