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The image you uploaded contains the function \( f(x) = \frac{1}{3}x^3 - 7 \), and you're asked to estimate the derivative at specific points: \( f'(4) \), \( f'(-5) \), and \( f'(9) \).

To estimate the derivative numerically, we can use the **difference quotient** formula:

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

Where \( h \) is a small interval value, and the smaller the value of \( h \), the more accurate the estimate of the derivative.

Let's calculate the derivatives at the required points \( x = 4 \), \( x = -5 \), and \( x = 9 \) using this method. I'll choose a small value for \( h \), such as \( h = 0.001 \).

I'll proceed with the calculations now.The estimated values of the derivatives are:

- \( f'(4) \approx 16.00 \)
- \( f'(-5) \approx 24.99 \)
- \( f'(9) \approx 81.01 \)

Would you like further details on any of these calculations or have any other questions?

Here are five related questions to deepen your understanding:

1. How does the value of \( h \) in the difference quotient affect the accuracy of the derivative estimate?
2. What is the exact derivative of the function \( f(x) = \frac{1}{3}x^3 - 7 \)?
3. Why is the derivative positive for both \( x = 4 \) and \( x = 9 \)?
4. How would you interpret the meaning of the derivative at \( x = -5 \)?
5. Can we use this approach to estimate derivatives for any type of function, or are there limitations?

**Tip:** The difference quotient method is great for approximating derivatives when you don't know the exact formula. However, it's essential to use a small enough \( h \) value to get an accurate estimate.

Let f(x) = 1/3 x^3 - 7. Estimate the derivative f'(4), f'(-5), and f'(9) using small enough intervals.

Learn how to estimate the derivatives of the function f(x) = 1/3 x^3 - 7 at points x = 4, x = -5, and x = 9 using the difference quotient method. This problem demonstrates a key calculus concept and is suitable for students in Grades 11-12 and college.

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