We will calculate the difference quotient:

$\frac{f(-1+h) - f(-1)}{h}$

for the function $f(x) = \frac{1}{x-1}$ and the specified values of $h$. Then, we will infer the integer $n$ such that the derivative matches $-\frac{1}{n^2}$.

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Step 1: Evaluate $f(-1)$

$f(-1) = \frac{1}{-1 - 1} = \frac{1}{-2} = -\frac{1}{2}.$

Step 2: General form of $f(-1 + h)$

$f(-1+h) = \frac{1}{(-1 + h) - 1} = \frac{1}{-2 + h}.$

Step 3: Difference Quotient

$\frac{f(-1+h) - f(-1)}{h} = \frac{\frac{1}{-2+h} - \left(-\frac{1}{2}\right)}{h} = \frac{\frac{1}{-2+h} + \frac{1}{2}}{h}.$

Simplify: $\frac{f(-1+h) - f(-1)}{h} = \frac{\frac{1}{-2+h} + \frac{1}{2}}{h} = \frac{\frac{2 + (-2 + h)}{(-2+h)(-2)}}{h} = \frac{\frac{h}{(-2+h)(-2)}}{h}.$

Cancel $h$ in the numerator and denominator: $\frac{f(-1+h) - f(-1)}{h} = \frac{1}{(-2+h)(-2)}.$

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Step 4: Compute for Specific Values of $h$

1. $h = 0.1$: $\frac{1}{(-2 + 0.1)(-2)} = \frac{1}{(-1.9)(-2)} = \frac{1}{3.8} \approx 0.2632.$

2. $h = 0.01$: $\frac{1}{(-2 + 0.01)(-2)} = \frac{1}{(-1.99)(-2)} = \frac{1}{3.98} \approx 0.2513.$

3. $h = -0.1$: $\frac{1}{(-2 - 0.1)(-2)} = \frac{1}{(-2.1)(-2)} = \frac{1}{4.2} \approx 0.2381.$

4. $h = -0.01$: $\frac{1}{(-2 - 0.01)(-2)} = \frac{1}{(-2.01)(-2)} = \frac{1}{4.02} \approx 0.2488.$

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Step 5: Derivative and Value of $n$

The derivative of $f(x) = \frac{1}{x-1}$ at $x = -1$ is computed as: $f'(x) = -\frac{1}{(x-1)^2}.$

At $x = -1$: $f'(-1) = -\frac{1}{(-2)^2} = -\frac{1}{4}.$

Given the form $-\frac{1}{n^2}$, it matches when $n = 2$.

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Final Answers:

1. $\frac{f(-1+h) - f(-1)}{h}$ for:

   • $h = 0.1$: $0.2632$
   • $h = 0.01$: $0.2513$
   • $h = -0.1$: $0.2381$
   • $h = -0.01$: $0.2488$

2. $n = 2$.

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Would you like me to expand on any of these steps? Here are 5 questions for further exploration:

1. How does the difference quotient approach the derivative as $h \to 0$?
2. Why does the derivative have the form $-\frac{1}{n^2}$?
3. How would this analysis change for a different function like $f(x) = \frac{1}{x+1}$?
4. Can we visualize these calculations on a graph of $f(x) = \frac{1}{x-1}$?
5. How do the values of $h$ influence the accuracy of the derivative approximation?

Tip: Always verify difference quotient computations by simplifying the algebraic expression before substituting values for $h$.