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We are given the function \( f(x) = \frac{3x - 2}{x + 3} \) and asked to find the value of \( f'(-2) \), the derivative of \( f(x) \) evaluated at \( x = -2 \).

### Step 1: Apply the Quotient Rule
The quotient rule is used when differentiating a function that is the ratio of two functions. For a function \( f(x) = \frac{g(x)}{h(x)} \), the quotient rule states:
\[
f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}
\]
For our function:
- \( g(x) = 3x - 2 \)
- \( h(x) = x + 3 \)

Now, we will differentiate \( g(x) \) and \( h(x) \):
- \( g'(x) = 3 \)
- \( h'(x) = 1 \)

### Step 2: Use the Quotient Rule
Now we can apply the quotient rule:
\[
f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}
\]
Substitute the values of \( g'(x) \), \( g(x) \), \( h'(x) \), and \( h(x) \):
\[
f'(x) = \frac{3(x + 3) - (3x - 2)(1)}{(x + 3)^2}
\]
Simplify the numerator:
\[
f'(x) = \frac{3(x + 3) - (3x - 2)}{(x + 3)^2}
\]
\[
f'(x) = \frac{3x + 9 - 3x + 2}{(x + 3)^2}
\]
\[
f'(x) = \frac{11}{(x + 3)^2}
\]

### Step 3: Evaluate at \( x = -2 \)
Now, substitute \( x = -2 \) into the derivative:
\[
f'(-2) = \frac{11}{(-2 + 3)^2}
\]
\[
f'(-2) = \frac{11}{1^2} = 11
\]

### Final Answer:
\[
f'(-2) = 11
\]

Would you like more details or have any other questions?

Here are 5 related questions to explore further:
1. How does the quotient rule differ from the product rule in differentiation?
2. What is the significance of the denominator in the quotient rule?
3. Can you explain why the derivative of a constant function is zero?
4. How would the answer change if the numerator had a higher degree polynomial?
5. How can you apply the chain rule to more complex functions?

**Tip:** The quotient rule is useful when differentiating functions where one term is divided by another. Keep practicing with different types of functions to get comfortable with this method.

If f(x) = (3x - 2) / (x + 3), then f'(-2) = ?

Find the derivative of f(x) = (3x - 2) / (x + 3) and evaluate f'(-2) using the quotient rule. This calculus problem involves differentiation techniques suitable for Grades 11-12.

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