1. **State the problem:** Determine the domain and range of the function $$y=\frac{1}{x-3}$$. 2. **Recall the domain rules:** The domain of a function is all the possible input values ($x$) for which the function is defined. Since division by zero is undefined, we must exclude values of $x$ that make the denominator zero. 3. **Find the domain:** Set the denominator equal to zero and solve: $$x-3=0$$ $$x=3$$ So, $x=3$ is not in the domain. Therefore, the domain is all real numbers except $3$, or in interval notation: $$(-\infty, 3) \cup (3, \infty)$$ 4. **Recall the range rules:** The range is all possible output values ($y$). For the function $$y=\frac{1}{x-3}$$, $y$ can take any real value except zero because the fraction can never equal zero (numerator is 1). 5. **Find the range:** Set $y=0$ and check if possible: $$0=\frac{1}{x-3}$$ This implies: $$1=0 \times (x-3)$$ $$1=0$$ which is false. So, $y=0$ is not in the range. Therefore, the range is all real numbers except $0$, or: $$(-\infty, 0) \cup (0, \infty)$$