Find domain y=square root (9x^6-x)/5x^3+9

We are tasked with finding the domain of the function:

\[
y = \frac{\sqrt{9x^6 - x}}{5x^3 + 9}
\]

### Step 1: Conditions for the square root
For the square root function to be defined, the expression inside the square root must be non-negative:

\[
9x^6 - x \geq 0
\]

Factor the left-hand side:

\[
x(9x^5 - 1) \geq 0
\]

This inequality holds true if both factors are either non-negative or non-positive. To solve it, let's analyze the signs of each factor:

1. \( x = 0 \) is a critical point.
2. \( 9x^5 - 1 = 0 \) gives us another critical point: \( x = \frac{1}{\sqrt[5]{9}} \).

Thus, we have to consider the sign of \( x(9x^5 - 1) \) in different intervals.

### Step 2: Denominator must not be zero
For the function to be defined, the denominator cannot be zero:

\[
5x^3 + 9 \neq 0
\]

Solve for \( x \):

\[
5x^3 \neq -9
\]

\[
x^3 \neq -\frac{9}{5}
\]

So, \( x \neq \sqrt[3]{-\frac{9}{5}} \), which is an additional restriction.

### Step 3: Combine restrictions
We combine the conditions from Steps 1 and 2 to determine the domain:

1. \( 9x^6 - x \geq 0 \)
2. \( 5x^3 + 9 \neq 0 \)

By solving these inequalities and removing the critical values, the domain of the function can be found as:

\[
\boxed{x \in (-\infty, -\sqrt[5]{9}] \cup (0, \infty)}
\]

### Would you like a more detailed explanation or any clarifications?

Here are 5 related questions:
1. What is the domain of \( y = \frac{\sqrt{x+2}}{x-3} \)?
2. How do you factor a polynomial expression like \( 9x^6 - x \)?
3. Can a function have a square root of a negative number?
4. How do critical points influence the domain of a function?
5. What happens if the denominator of a function equals zero?

**Tip:** Always remember to check for both square root and denominator restrictions when finding the domain of a function.

Find the domain of the function y = sqrt(9x^6 - x) / (5x^3 + 9)

Discover how to find the domain of the function y = sqrt(9x^6 - x) / (5x^3 + 9) by analyzing the square root conditions and the restrictions on the denominator. This detailed solution walks through factoring, solving inequalities, and critical points. Perfect for Grades 10-12 students learning about functions and domain.

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