3) State the domain using interval notation. a) $\frac{8 x-5}{x^{2}-25}$ b) $\frac{x^{3}+8 x^{4}}{x^{2}+25}$ c) $\sqrt{x^{2}-5 x-6}$

Step 1 ：The domain of a function is the set of all possible input values (often the 'x' variable), which produce a valid output from a particular function. The domain can be found by identifying the values of x that can be plugged into the function without violating any mathematical rules or constraints.

Step 2 ：For the function \(\frac{8 x-5}{x^{2}-25}\), the denominator cannot be zero because division by zero is undefined in mathematics. Therefore, we need to find the values of x that make the denominator zero and exclude them from the domain. The domain of the first function is all real numbers except -5 and 5. In interval notation, this is \((-\infty, -5) \cup (-5, 5) \cup (5, \infty)\).

Step 3 ：For the function \(\frac{x^{3}+8 x^{4}}{x^{2}+25}\), the denominator can never be zero because the square of any real number is always non-negative and we are adding 25 to it. Therefore, the domain is all real numbers. In interval notation, this is \((-\infty, \infty)\).

Step 4 ：For the function \(\sqrt{x^{2}-5 x-6}\), the expression under the square root must be non-negative because the square root of a negative number is not a real number. Therefore, we need to find the values of x that make the expression under the square root non-negative. The domain of the third function is all x such that \(x^2 - 5x - 6 \geq 0\). We can solve this inequality to find the interval. The domain of the third function is all x such that \(x \leq -1\) or \(x \geq 6\). In interval notation, this is \((-\infty, -1] \cup [6, \infty)\).

Step 5 ：Final Answer: \(\boxed{\text{a) The domain of the function } \frac{8 x-5}{x^{2}-25} \text{ is } (-\infty, -5) \cup (-5, 5) \cup (5, \infty)}\)

Step 6 ：\(\boxed{\text{b) The domain of the function } \frac{x^{3}+8 x^{4}}{x^{2}+25} \text{ is } (-\infty, \infty)}\)

Step 7 ：\(\boxed{\text{c) The domain of the function } \sqrt{x^{2}-5 x-6} \text{ is } (-\infty, -1] \cup [6, \infty)}\)