The question shown in the image asks to find the domain and range of the function provided:

$f(x) = \frac{5}{x - 3}$

1. Domain of $f(x)$:

The domain of a function is the set of all possible input values (i.e., $x$) for which the function is defined. Since the function $f(x) = \frac{5}{x - 3}$ involves a denominator, the function will be undefined when the denominator is zero.

• Set the denominator equal to zero and solve for $x$: $x - 3 = 0$ $x = 3$ Thus, the function is undefined at $x = 3$. Therefore, the domain of the function is all real numbers except $x = 3$.

The domain in interval notation is: $(-\infty, 3) \cup (3, +\infty)$

2. Range of $f(x)$:

The range of a function is the set of all possible output values (i.e., $f(x)$).

Since $f(x) = \frac{5}{x - 3}$, we know that the function can never equal zero because the numerator (5) is a constant, and the denominator will never cause the expression to be zero. Thus, the output can be any real number except zero.

Therefore, the range of the function is all real numbers except $f(x) = 0$.

The range in interval notation is: $(-\infty, 0) \cup (0, +\infty)$

Final Answer:

• Domain: $(-\infty, 3) \cup (3, +\infty)$
• Range: $(-\infty, 0) \cup (0, +\infty)$

Would you like more details or have any questions?

Here are five related questions:

1. How do you determine the domain of a function involving square roots?
2. What happens to the range if the numerator of a rational function is zero?
3. Can the function $f(x) = \frac{5}{x - 3}$ ever have a vertical asymptote? Explain.
4. What is the horizontal asymptote of the function $f(x) = \frac{5}{x - 3}$?
5. How would the domain and range change if the function was $f(x) = \frac{5}{(x - 3)^2}$?

Tip: When determining the domain, always look for values that make the denominator zero or cause square roots to be negative.