how does the denominator being in a square root change things

When the denominator of a function includes a square root, it introduces additional constraints to ensure the function is defined. Specifically:

1. **Non-Negativity Condition**: The expression inside the square root must be non-negative because the square root of a negative number is not a real number. So, for \(\sqrt{x^2 + x - 2}\) to be defined, we need:
   \[
   x^2 + x - 2 \geq 0
   \]

2. **Non-Zero Condition**: Since the expression is in the denominator, it must not be zero because division by zero is undefined. Therefore, we also need:
   \[
   x^2 + x - 2 \neq 0
   \]

These conditions together define the domain of the function.

Let's revisit the problem step-by-step, keeping these conditions in mind:

### Step 1: Non-Negativity Condition

We already factored the quadratic expression:
\[ x^2 + x - 2 = (x + 2)(x - 1) \]

To find where \( (x + 2)(x - 1) \geq 0 \), we analyze the intervals around the critical points \( x = -2 \) and \( x = 1 \):

- For \( x \in (-\infty, -2) \):
  \[
  (x + 2) < 0 \quad \text{and} \quad (x - 1) < 0 \quad \Rightarrow \quad (x + 2)(x - 1) > 0
  \]

- For \( x \in (-2, 1) \):
  \[
  (x + 2) > 0 \quad \text{and} \quad (x - 1) < 0 \quad \Rightarrow \quad (x + 2)(x - 1) < 0
  \]

- For \( x \in (1, \infty) \):
  \[
  (x + 2) > 0 \quad \text{and} \quad (x - 1) > 0 \quad \Rightarrow \quad (x + 2)(x - 1) > 0
  \]

Thus, the non-negativity condition is satisfied for:
\[ x \in (-\infty, -2] \cup [1, \infty) \]

### Step 2: Non-Zero Condition

We exclude the points where the expression inside the square root equals zero:
\[ x^2 + x - 2 = 0 \]
This occurs at \( x = -2 \) and \( x = 1 \).

### Combining the Conditions

We combine the intervals where \( x^2 + x - 2 \geq 0 \) and exclude the points where \( x^2 + x - 2 = 0 \):
\[ \boxed{(-\infty, -2) \cup (-2, 1) \cup (1, \infty)} \]

In summary, the square root in the denominator imposes two key restrictions: the expression inside the square root must be non-negative, and it must not be zero. This ensures the function is both real-valued and defined.

Would you like further details or have any questions?

Here are 8 related questions to deepen your understanding:

1. How does the presence of a square root in the numerator affect the domain?
2. Can a square root ever be negative for real numbers?
3. What changes if the square root expression is in the numerator instead of the denominator?
4. How do you graphically represent the solution to \( (x + 2)(x - 1) \geq 0 \)?
5. What is the domain of \( g(x) = \frac{1}{\sqrt{x-3}} \)?
6. How do you handle square root functions that include absolute values?
7. What happens to the domain if the expression inside the square root is always positive?
8. How do you determine the domain of composite functions involving square roots and polynomials?

**Tip:** Always carefully analyze both the non-negativity and non-zero conditions when dealing with square roots in the denominator to accurately determine the domain.

Explore how square roots in denominators affect function domains with detailed explanations and examples. Learn about non-negativity and non-zero conditions crucial for defining functions involving square roots in denominators. Suitable for students in Grades 10-12.

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