Let's work through the problem step by step.

Given function:

$f(x) = \sqrt[3]{256x^3 + 256}$

Step 1: Find the equation of the tangent line to the graph of the function at $x = 1$.

To find the equation of the tangent line, we need to use the formula for the equation of a line: $L(x) = f'(x_0)(x - x_0) + f(x_0)$ where $x_0 = 1$, $f'(x)$ is the derivative of the function, and $f(x_0)$ is the function value at $x = 1$.

1. Calculate $f(1)$: [ f(1) = \sqrt[3]{256(1)^3 + 256} = \sqrt[3]{512} = 8 ]

2. Find $f'(x)$ (the derivative of the function):

   • Apply the chain rule to differentiate $f(x) = \sqrt[3]{256x^3 + 256}$.

$f'(x) = \frac{d}{dx}\left( (256x^3 + 256)^{1/3} \right) = \frac{1}{3}(256x^3 + 256)^{-2/3} \cdot (768x^2)$ $f'(x) = \frac{768x^2}{3(256x^3 + 256)^{2/3}} = \frac{256x^2}{(256x^3 + 256)^{2/3}}$

3. Evaluate $f'(1)$: $f'(1) = \frac{256(1)^2}{(256(1)^3 + 256)^{2/3}} = \frac{256}{512^{2/3}} = \frac{256}{64} = 4$

Now that we have $f(1) = 8$ and $f'(1) = 4$, the equation of the tangent line is: $L(x) = 4(x - 1) + 8$ Simplifying: $L(x) = 4x - 4 + 8 = 4x + 4$

Step 2: Use the tangent line to approximate $f(1.1)$.

Using the equation of the tangent line $L(x) = 4x + 4$, we can approximate $f(1.1)$ by calculating $L(1.1)$: $L(1.1) = 4(1.1) + 4 = 4.4 + 4 = 8.4$

Step 3: Compute the actual value of $f(1.1)$ and find the error.

1. Actual value of $f(1.1)$: [ f(1.1) = \sqrt[3]{256(1.1)^3 + 256} = \sqrt[3]{256(1.331) + 256} = \sqrt[3]{340.736 + 256} = \sqrt[3]{596.736} \approx 8.37866 ]

2. Calculate the error: $\text{Error} = |f(1.1) - L(1.1)| = |8.37866 - 8.4| = 0.02134$

Final Answers:

• Equation of the tangent line: $L(x) = 4x + 4$
• Approximation of $f(1.1)$ using the tangent line: $L(1.1) = 8.4$
• Actual value of $f(1.1)$: $f(1.1) \approx 8.37866$
• Error: $\approx 0.02134$

Do you need further details or explanations? Here are some follow-up questions you might find useful:

1. How is the chain rule applied when differentiating functions involving cube roots?
2. What is the significance of the tangent line in approximating values of the function?
3. How would the tangent line approximation change if the function's curvature was different at $x = 1$?
4. Can the error between the approximation and the actual value be reduced, and how?
5. What other methods could be used to approximate $f(1.1)$ besides using a tangent line?

Tip: For better accuracy in approximations, the second derivative (concavity) can sometimes be considered, leading to a quadratic approximation.