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The given expression asks for the difference quotient, which is a form of the derivative definition, for the function \( f(x) = \frac{1}{x} \) at \( x = 2 \).

We are asked to evaluate:

\[
\frac{f(2+h) - f(2)}{h}
\]

### Step-by-step solution:

1. **Substitute \( f(x) = \frac{1}{x} \) into the expression:**
   \[
   f(2+h) = \frac{1}{2+h}, \quad f(2) = \frac{1}{2}
   \]
   
2. **Rewrite the difference quotient:**
   \[
   \frac{f(2+h) - f(2)}{h} = \frac{\frac{1}{2+h} - \frac{1}{2}}{h}
   \]

3. **Simplify the numerator:**
   To combine the two fractions, we need a common denominator:
   \[
   \frac{1}{2+h} - \frac{1}{2} = \frac{2 - (2+h)}{(2+h)(2)} = \frac{-h}{2(2+h)}
   \]

4. **Substitute this back into the difference quotient:**
   \[
   \frac{\frac{-h}{2(2+h)}}{h} = \frac{-h}{h \cdot 2(2+h)} = \frac{-1}{2(2+h)}
   \]

5. **As \( h \to 0 \), the expression becomes:**
   \[
   \frac{-1}{2(2+0)} = \frac{-1}{4}
   \]

Thus, the value of the difference quotient is \( \frac{-1}{4} \).

---

Would you like further details or clarifications on any step? 

Here are some related questions you might find useful:

1. What is the general form of the difference quotient?
2. How does the difference quotient relate to the derivative?
3. What is the derivative of \( f(x) = \frac{1}{x} \)?
4. Why does the limit as \( h \to 0 \) give the derivative?
5. How can this process be applied to other functions?

**Tip:** The difference quotient is foundational to understanding how derivatives work, as it gives an approximation of the rate of change over small intervals.

If f(x) = 1/x, then (f(2+h) - f(2)) / h equals?

Learn how to compute the difference quotient for the function f(x) = 1/x at x = 2. This detailed solution involves calculus concepts such as the difference quotient and limits, suitable for students in Grades 11-12.

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