Let's start by analyzing the given function in the image.

The function to examine is:

$f(x) = \frac{\sqrt{5x - 10}}{x^2 - 36}$

Step 1: Domain

The domain of a function refers to the values of $x$ for which the function is defined.

1. The square root in the numerator:

The expression inside the square root must be non-negative for real values, so: $5x - 10 \geq 0$ Solving for $x$: $5x \geq 10 \quad \Rightarrow \quad x \geq 2$

2. The denominator:

The denominator cannot be zero, so: $x^2 - 36 \neq 0$ Factor the denominator: $(x - 6)(x + 6) \neq 0$ Thus, $x \neq 6$ and $x \neq -6$.

Domain Conclusion:

The domain is $x \geq 2$ but excluding $x = 6$. Therefore, the domain in interval notation is: $[2, 6) \cup (6, \infty)$

Step 2: Range

To find the range, we need to consider the behavior of the function. The square root function always gives non-negative values, so the numerator $\sqrt{5x - 10}$ is non-negative.

• As $x \to 2^+$, the numerator approaches zero and the denominator approaches 0 as well, but from the positive side. So, the function approaches $+\infty$.
• As $x \to \infty$, both the numerator and denominator grow large, but the denominator grows faster than the numerator, meaning the function approaches 0.

Range Conclusion:

The range is all non-negative values, but it never actually reaches 0. Therefore, the range is: $[0, \infty)$

Let me know if you need any further clarifications!

Further questions:

1. What is the behavior of the function near $x = 6$?
2. How do transformations affect the domain and range of functions like this one?
3. Can this function be simplified or rewritten in a different form for easier analysis?
4. What happens to the function if we change the square root expression to something else, like $\sqrt{x + 2}$?
5. How would you graph this function?

Tip: When dealing with square roots and fractions, always start by checking the restrictions on the domain (i.e., where the expressions inside the root are positive and where the denominator is nonzero).