To find the domain of the function

$g(x) = \frac{\sqrt{x}}{6x^2 + 5x - 1}$

we need to analyze two main components of the function:

1. The square root in the numerator, $\sqrt{x}$, must have a non-negative argument, so $x \geq 0$.
2. The denominator, $6x^2 + 5x - 1$, must not be zero because division by zero is undefined.

Step 1: Restriction from the square root

For $\sqrt{x}$ to be defined, we require: $x \geq 0$ This gives the preliminary condition that the domain is restricted to $x \geq 0$.

Step 2: Restriction from the denominator

Next, we need to ensure the denominator is not zero. The denominator is: $6x^2 + 5x - 1 = 0$ Solving this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 6$, $b = 5$, and $c = -1$. Substituting into the formula: $x = \frac{-5 \pm \sqrt{5^2 - 4(6)(-1)}}{2(6)}$ $x = \frac{-5 \pm \sqrt{25 + 24}}{12}$ $x = \frac{-5 \pm \sqrt{49}}{12}$ $x = \frac{-5 \pm 7}{12}$ Thus, the two solutions are: $x = \frac{-5 + 7}{12} = \frac{2}{12} = \frac{1}{6}$ $x = \frac{-5 - 7}{12} = \frac{-12}{12} = -1$

The denominator is zero when $x = \frac{1}{6}$ and $x = -1$. Since $x = -1$ is excluded from the domain because of the square root, we only need to exclude $x = \frac{1}{6}$.

Step 3: Combine the conditions

The function is defined for $x \geq 0$, except at $x = \frac{1}{6}$. Therefore, the domain is:

$[0, \frac{1}{6}) \cup (\frac{1}{6}, \infty)$

This is the interval notation solution provided.

Would you like further details on any part of the process? Here are some related questions:

1. How do you solve quadratic equations in general?
2. What is the significance of excluding $\frac{1}{6}$ from the domain?
3. Why must $x \geq 0$ for the square root function to be defined?
4. How does the quadratic formula work in this context?
5. What other techniques can be used to find the domain of a function?

Tip: Always check both the numerator and denominator when finding the domain of a rational function.