Let's analyze the function $f(x) = \frac{1}{\sqrt{x - 3}}$ to determine its domain and range.

1. Domain:

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

• The expression under the square root, $x - 3$, must be non-negative because the square root of a negative number is not real. Therefore, we need: $x - 3 \geq 0$ $x \geq 3$

• Additionally, the function $f(x)$ involves a fraction, and the denominator $\sqrt{x - 3}$ cannot be zero because division by zero is undefined. Therefore, we require: $\sqrt{x - 3} \neq 0$ $x - 3 \neq 0$ $x \neq 3$

Thus, combining these two conditions, the domain is: $\text{Domain of } f(x): \, (3, \infty)$ This means $f(x)$ is defined for all $x > 3$.

2. Range:

The range of a function is the set of all possible output values (y-values).

• $f(x) = \frac{1}{\sqrt{x - 3}}$ is positive because both the numerator (1) and the square root function are positive for $x > 3$.

• As $x$ approaches $3$ from the right (i.e., $x$ is slightly greater than 3), $\sqrt{x - 3}$ approaches 0, and therefore, $f(x)$ becomes very large. Thus, the function can take very large positive values.

• As $x$ increases, $\sqrt{x - 3}$ increases, making $f(x)$ decrease. However, since the square root function increases without bound as $x$ increases, $f(x)$ approaches 0 but never actually reaches 0.

Thus, the range of the function is: $\text{Range of } f(x): \, (0, \infty)$

Summary:

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

Would you like more details or have any questions?

Here are some related questions to consider:

1. How would the domain change if the function were $f(x) = \frac{1}{\sqrt{5 - x}}$?
2. What is the domain and range of $f(x) = \sqrt{x - 2}$?
3. How does adding a constant inside the square root affect the domain and range?
4. What is the domain and range of $f(x) = \frac{1}{\sqrt{x^2 - 1}}$?
5. How would the range change if the function were $f(x) = \frac{2}{\sqrt{x - 3}}$?

Tip: Always remember that square roots require the radicand to be non-negative, and denominators cannot be zero!