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To find the domain of the function 

\[
f(x) = \sqrt{\frac{x+3}{x-2}}
\]

we need to ensure two things:

1. The expression inside the square root must be non-negative because we cannot take the square root of a negative number (in real numbers).
   
   This gives us the condition:
   \[
   \frac{x+3}{x-2} \geq 0
   \]

2. The denominator \(x - 2\) cannot be zero because division by zero is undefined.

Let's solve these step by step:

### Step 1: Solve \(\frac{x+3}{x-2} \geq 0\)

To solve the inequality \(\frac{x+3}{x-2} \geq 0\), we need to find where the expression is either zero or changes sign. The expression \(\frac{x+3}{x-2}\) will be zero when the numerator is zero:

\[
x + 3 = 0 \implies x = -3
\]

The expression will be undefined when the denominator is zero:

\[
x - 2 = 0 \implies x = 2
\]

So the critical points are \(x = -3\) (where the expression is zero) and \(x = 2\) (where the expression is undefined).

### Step 2: Test intervals

Now we test intervals between these critical points to see where the expression is positive or negative. The intervals are:
1. \(x < -3\)
2. \(-3 < x < 2\)
3. \(x > 2\)

- For \(x < -3\) (e.g., \(x = -4\)): 
  \[
  \frac{-4 + 3}{-4 - 2} = \frac{-1}{-6} = \frac{1}{6} > 0
  \]
  This interval is positive.

- For \(-3 < x < 2\) (e.g., \(x = 0\)): 
  \[
  \frac{0 + 3}{0 - 2} = \frac{3}{-2} = -\frac{3}{2} < 0
  \]
  This interval is negative.

- For \(x > 2\) (e.g., \(x = 3\)): 
  \[
  \frac{3 + 3}{3 - 2} = \frac{6}{1} = 6 > 0
  \]
  This interval is positive.

### Step 3: Combine results

- The expression is non-negative when \(x \in (-\infty, -3] \cup (2, \infty)\).
- We exclude \(x = 2\) because the function is undefined there.

### Final Answer: Domain of \(f(x)\)

\[
\boxed{(-\infty, -3] \cup (2, \infty)}
\]

Would you like more details on this solution or clarification?

Here are some related questions:
1. How do you determine when a rational function is positive or negative?
2. Can you explain the concept of critical points in inequalities?
3. Why is division by zero undefined in mathematical functions?
4. How would you solve a similar problem with an absolute value instead of a square root?
5. What happens if we change the square root function to a cube root?

**Tip:** Always check both the numerator and the denominator when solving inequalities with fractions, as both affect the sign of the expression.

Find the domain of the function f(x) = √((x + 3)/(x - 2))

Learn how to find the domain of the function f(x) = √((x + 3)/(x - 2)) by solving inequalities and identifying where the rational function is non-negative. Suitable for high school students in Grades 9-12.

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