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The question you are referring to is number 31, which states:

**31. \( f(x) = \frac{1}{x^2} \)**

You are likely being asked to express \(\frac{f(x+h) - f(x)}{h}\) as a single simplified expression.

### Solution:

1. **Find \( f(x+h) \):**

   \[
   f(x+h) = \frac{1}{(x+h)^2}
   \]

2. **Form the difference quotient:**

   \[
   \frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}
   \]

3. **Simplify the numerator:**

   To simplify the expression, combine the fractions in the numerator:

   \[
   \frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2(x+h)^2}
   \]

   Expand the squared term:

   \[
   (x+h)^2 = x^2 + 2xh + h^2
   \]

   So the numerator becomes:

   \[
   x^2 - (x^2 + 2xh + h^2) = x^2 - x^2 - 2xh - h^2 = -2xh - h^2
   \]

   Therefore:

   \[
   \frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{-2xh - h^2}{x^2(x+h)^2}
   \]

4. **Divide by \( h \):**

   Now, divide the entire expression by \( h \):

   \[
   \frac{-2xh - h^2}{hx^2(x+h)^2} = \frac{-2x - h}{x^2(x+h)^2}
   \]

So, the simplified form of \(\frac{f(x+h) - f(x)}{h}\) is:

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

This is the final simplified expression.

Would you like more details on any part of the process or have any questions? Here are five related questions that could help deepen your understanding:

1. How does this expression relate to the concept of the derivative?
2. What happens to the expression as \(h\) approaches 0?
3. How would you compute the derivative \(f'(x)\) directly from the original function \(f(x)\)?
4. How does this process generalize to other functions \(f(x)\)?
5. Can you simplify \(\frac{-2x - h}{x^2(x+h)^2}\) further when \(h\) is very small?

**Tip:** When simplifying expressions, always consider factoring or expanding polynomials before combining terms. This approach often leads to more straightforward simplifications.

Explore the difference quotient problem involving the function f(x) = 1/x^2. Learn how to simplify and solve the difference quotient expression step-by-step. This problem delves into calculus concepts such as limits and the definition of derivatives, suitable for advanced high school or college students.

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