Let's find the difference quotient for the function \( f(x) = \frac{1}{x+3} \).

The difference quotient is given by:

\[
\frac{f(x+h) - f(x)}{h}
\]

### 1. **Calculate \( f(x+h) \):**

\[
f(x+h) = \frac{1}{(x+h)+3} = \frac{1}{x+h+3}
\]

### 2. **Find the difference \( f(x+h) - f(x) \):**

\[
f(x+h) - f(x) = \frac{1}{x+h+3} - \frac{1}{x+3}
\]

To subtract these fractions, we need a common denominator:

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

Simplify the numerator:

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

### 3. **Form the difference quotient:**

\[
\frac{f(x+h) - f(x)}{h} = \frac{-h}{h \cdot (x+h+3)(x+3)}
\]

Simplify by canceling \( h \) in the numerator and denominator:

\[
\frac{f(x+h) - f(x)}{h} = \frac{-1}{(x+h+3)(x+3)}
\]

So, the difference quotient for \( f(x) = \frac{1}{x+3} \) is:

\[
\frac{f(x+h) - f(x)}{h} = \frac{-1}{(x+h+3)(x+3)}
\]

Would you like further clarification or have any questions?

---

Here are 5 related questions for further exploration:

1. How does the difference quotient change as \( h \) approaches 0 for \( f(x) = \frac{1}{x+3} \)?
2. What is the derivative of \( f(x) = \frac{1}{x+3} \) using the definition of the derivative?
3. How would the difference quotient change for a function with a different denominator, such as \( f(x) = \frac{1}{x+5} \)?
4. Can you apply the difference quotient to a function like \( f(x) = \frac{2}{x+3} \)?
5. How does the concept of limits apply to the difference quotient?

**Tip:** When working with rational functions, always remember to find a common denominator when subtracting fractions. This simplifies the difference quotient and prepares it for taking the limit as \( h \) approaches 0.

Explore how to find the difference quotient for the function f(x) = 1/(x+3). This detailed explanation covers step-by-step calculations and insights into calculus concepts such as difference quotients and limits. Suitable for advanced high school students and calculus enthusiasts.

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