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The image you uploaded contains mathematical problems related to **secant slopes**, **derivatives**, and **tangent-line approximations**. Let’s walk through the key parts of the exercise.

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### Summary of Concepts:
1. **Secant Slope / Difference Quotient**:
   - **Forward Difference Quotient**:  
     \[ \frac{f(a + \Delta x) - f(a)}{\Delta x} \]  
   This approximates the derivative \( f'(a) \) using a small change \( \Delta x \) from the point \( a \).
   
   - **Centered Difference Quotient**:  
     \[ \frac{f(a + \Delta x) - f(a - \Delta x)}{2\Delta x} \]  
   This is a more refined approximation of \( f'(a) \), using points on both sides of \( a \).

2. **Problems to Solve**:
   - **Problem 1**: Compute the derivative of \( f(x) = x^3 + 2x + 1 \) at \( a = 1 \).  
     - **Hint**: Use the **derivative rules** and apply the given identity:
       \[ b^3 - a^3 = (b - a)(b^2 + ab + a^2). \]

   - **Problem 2**: Explore the function \( f(x) = \sqrt[3]{x} = x^{1/3} \).  
     You are asked to:
     1. Compute \( f'(8) \).
     2. Use a **tangent-line approximation** at \( a = 8 \) to estimate values like \( \sqrt[3]{8.2} \) and \( \sqrt[3]{7.7} \).
     3. Compare your approximation to the exact values using MATLAB or a calculator.
     4. Plot the graph of \( f(x) \) and its tangent at \( a = 8 \) and interpret the results visually.

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### Step-by-Step Solution Ideas:
#### **Problem 1**: Derivative of \( f(x) = x^3 + 2x + 1 \) at \( a = 1 \).
1. Find \( f'(x) \) using standard derivative rules:
   \[ f'(x) = 3x^2 + 2. \]
2. Now, evaluate the derivative at \( a = 1 \):
   \[ f'(1) = 3(1)^2 + 2 = 5. \]

#### **Problem 2**: Exploring \( f(x) = \sqrt[3]{x} = x^{1/3} \).
(a) **Compute \( f'(8) \)**:  
   Use the **power rule** for derivatives:
   \[ f'(x) = \frac{1}{3}x^{-2/3}. \]  
   Now, substitute \( x = 8 \):
   \[ f'(8) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}. \]

(b) **Linear Approximation**:
   - The equation of the tangent line at \( a = 8 \) is:
     \[ y - f(8) = f'(8)(x - 8). \]
   - Since \( f(8) = \sqrt[3]{8} = 2 \) and \( f'(8) = \frac{1}{12} \), the tangent line is:
     \[ y - 2 = \frac{1}{12}(x - 8). \]
     Rearranging:
     \[ y = \frac{1}{12}x + \frac{8}{12} = \frac{1}{12}x + \frac{2}{3}. \]

(c) **Estimate \( \sqrt[3]{8.2} \) and \( \sqrt[3]{7.7} \)**:
   - Plug \( x = 8.2 \) and \( x = 7.7 \) into the tangent line equation to get the estimates.
   - For example, for \( x = 8.2 \):
     \[ y = \frac{1}{12}(8.2) + \frac{2}{3}. \]

(d) **Graphing**:
   - Plot \( f(x) = \sqrt[3]{x} \) and the tangent line to visualize the approximation at \( a = 8 \).

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Do you want a detailed solution for any part of these problems, such as plotting guidance or MATLAB hints?

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### 5 Related Questions:
1. What is the significance of the difference quotient in estimating derivatives?
2. How does a centered difference quotient differ from a forward difference quotient in terms of accuracy?
3. Can you derive the general formula for a tangent-line approximation at any point \( a \)?
4. How would the tangent-line approximation behave for a rapidly changing function?
5. What other numerical methods exist for approximating derivatives besides the difference quotient?

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**Tip**: Always double-check your derivative approximations by graphing both the function and its tangent line!

Some of the book problems use terminology for secant slopes. Let f be a function defined on an open interval containing the number a and Δx ≠ 0. Using difference quotients, compute the derivative of the function f(x) = x^3 + 2x + 1 at a = 1 and explore the function f(x) = ∛x at a = 8.

Learn how to compute the derivative of f(x) = x^3 + 2x + 1 at a = 1 using difference quotients, including forward and centered methods. Explore the function f(x) = ∛x at a = 8, use tangent line approximations, and visualize the results. Ideal for high school students in grades 11-12 or college calculus.

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