jmc

Solution #1

To determine the range, we suppose $y=\frac{3x+1}{x+8}$ (where $x\ne -8$) and see if we can solve for $x$: \begin{array}{r r@{~=~}l} & y & (3x+1)/(x+8) \\ \Leftrightarrow & y(x + 8) & 3x + 1 \\ \Leftrightarrow & yx + 8y & 3x + 1 \\ \Leftrightarrow & x(y - 3) & 1 - 8y. \end{array}This last equation gives a contradiction if $y=3$, since in this case it says that $0=-23$. Therefore, it is impossible for $g(x)$ to equal $3$ for any value of $x$. But for any value of $y$ other than $3$, the last equation can be solved to yield $x=\frac{1-8y}{y-3}$, or, in other words, $g\left(\frac{1-8y}{y-3}\right)=y$.

Therefore, the range of $g(x)$ is $\mathbb{R}\setminus \{3\}=\boxed{(-\infty ,3)\cup (3,\infty )}$.

Solution #2

We can rewrite $g(x)$ as follows: $$g(x)=\frac{3x+1}{x+8}=\frac{3x+24}{x+8}-\frac{23}{x+8}=3-\frac{23}{x+8}.$$Then we note that $x+8$ takes on all real values, so $\frac{1}{x+8}$ takes on every value which is the reciprocal of some nonzero real number, i.e., $\frac{1}{x+8}$ takes on all nonzero values. Accordingly, $3-\frac{23}{x+8}$ takes on all values not equal to $3$.

Therefore, the range of $g(x)$ is $\mathbb{R}\setminus \{3\}=\boxed{(-\infty ,3)\cup (3,\infty )}$.