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The image shows a problem where the rate of change of the function \( f(x) = x^2 - x \) is being estimated at \( x = -1 \) using the difference quotient:

\[
\frac{f(-0.999) - f(-1)}{-0.999 - (-1)} = -2.999
\]

This is an approximation of the derivative of \( f(x) \) at \( x = -1 \), which is the slope of the tangent line at that point. To break this down:

1. **Function given**: \( f(x) = x^2 - x \)
2. **Points considered**: \( x = -1 \) and \( x = -0.999 \)
3. **Difference quotient formula**:
   \[
   \frac{f(x_1) - f(x_0)}{x_1 - x_0}
   \]
   where \( x_1 = -0.999 \) and \( x_0 = -1 \).

### Steps:
1. Calculate \( f(-1) \):
   \[
   f(-1) = (-1)^2 - (-1) = 1 + 1 = 2
   \]
2. Calculate \( f(-0.999) \):
   \[
   f(-0.999) = (-0.999)^2 - (-0.999) = 0.998001 + 0.999 = 1.997001
   \]
3. Apply the difference quotient:
   \[
   \frac{f(-0.999) - f(-1)}{-0.999 - (-1)} = \frac{1.997001 - 2}{-0.999 + 1} = \frac{-0.002999}{0.001} = -2.999
   \]

Thus, the result of the difference quotient is \( -2.999 \), which approximates the derivative at \( x = -1 \).

Would you like more details on the process, or have any specific questions?

Here are 5 relative questions:
1. How does the difference quotient relate to the derivative of a function?
2. Can this method be used for any function, or only for specific types?
3. How would the result change if we used \( x = -0.99 \) instead of \( x = -0.999 \)?
4. What is the exact derivative of \( f(x) = x^2 - x \) at \( x = -1 \)?
5. Why is it important to use values very close to \( x = -1 \) for this approximation?

**Tip**: The difference quotient becomes more accurate as the points used in the calculation get closer to each other.

Estimate the rate of change of f(x) = x^2 - x at x = -1 using the difference quotient.

Learn how to estimate the rate of change of the function f(x) = x^2 - x at x = -1 using the difference quotient method. This example breaks down the steps and provides an approximation of the derivative as -2.999, suitable for students studying calculus.

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