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.

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.

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 $a = 8$.

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

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?

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