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The problem asks us to evaluate the difference quotient for the given function \( f(x) = \frac{1}{x+1} \). The difference quotient is the expression:

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

### Step-by-step solution:

1. **Substitute \( f(x) = \frac{1}{x+1} \) and \( f(x+h) \) into the difference quotient formula:**

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

   So the difference quotient becomes:

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

2. **Combine the fractions in the numerator:**

   We need a common denominator to subtract these two fractions. The common denominator will be \( (x+h+1)(x+1) \). So:

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

3. **Simplify the numerator:**

   The numerator simplifies as follows:

   \[
   (x+1) - (x+h+1) = x+1 - x - h - 1 = -h
   \]

   So we now have:

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

4. **Substitute back into the difference quotient:**

   Now, the expression for the difference quotient becomes:

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

5. **Simplify the expression:**

   Dividing by \( h \) is the same as multiplying by \( \frac{1}{h} \), so:

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

### Final result:
\[
\frac{f(x+h) - f(x)}{h} = \frac{-1}{(x+h+1)(x+1)}
\]

This is the simplified form of the difference quotient for the function \( f(x) = \frac{1}{x+1} \).

Would you like further details on any step or have any other questions?

Here are 5 related questions:
1. How does this difference quotient relate to the derivative of the function?
2. Can we simplify the difference quotient further when \( h \to 0 \)?
3. What is the geometric interpretation of the difference quotient?
4. How do you compute the derivative of a rational function like \( f(x) = \frac{1}{x+1} \)?
5. How does this method apply to other functions of similar forms?

**Tip:** The difference quotient is a fundamental concept in calculus as it forms the basis for the definition of the derivative.

If f(x) = 1/(x+1), evaluate the difference quotient (f(x+h) - f(x))/h.

Explore the step-by-step evaluation of the difference quotient for the rational function f(x) = 1/(x+1). This problem introduces key calculus concepts, including the difference quotient and its simplification. Suitable for high school students in grades 11-12 or college calculus courses.

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