Let's go through the problem step by step. Here’s the breakdown:

Problem

The function is given as: $f(x) = \sqrt[3]{243x^3 + 486}$

Task 1: Find the Equation of the Tangent Line at $x = 1$

To find the tangent line $L(x)$ at $x = 1$, we need:

1. The value of $f(1)$.
2. The derivative $f'(x)$ at $x = 1$, which gives the slope of the tangent.

Step 1.1: Compute $f(1)$

Substitute $x = 1$ into $f(x)$: [ f(1) = \sqrt[3]{243(1)^3 + 486} = \sqrt[3]{243 + 486} = \sqrt[3]{729} = 9 ]

So, $f(1) = 9$.

Step 1.2: Compute $f'(x)$ Using the Chain Rule

To find $f'(x)$, use the chain rule. Rewrite $f(x)$ as: $f(x) = (243x^3 + 486)^{1/3}$

Let $g(x) = 243x^3 + 486$ and $h(x) = g(x)^{1/3}$, so that $f(x) = h(g(x))$.

1. Differentiate $h(x) = x^{1/3}$: $h'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}}$
2. Differentiate $g(x) = 243x^3 + 486$: $g'(x) = 729x^2$

Using the chain rule, $f'(x) = h'(g(x)) \cdot g'(x)$: $f'(x) = \frac{1}{3(243x^3 + 486)^{2/3}} \cdot 729x^2$

Simplify this expression: $f'(x) = \frac{729x^2}{3(243x^3 + 486)^{2/3}} = \frac{243x^2}{(243x^3 + 486)^{2/3}}$

Step 1.3: Compute $f'(1)$

Substitute $x = 1$ into $f'(x)$: $f'(1) = \frac{243 \cdot 1^2}{(243 \cdot 1^3 + 486)^{2/3}} = \frac{243}{729^{2/3}}$

Since $729^{2/3} = 81$, we get: $f'(1) = \frac{243}{81} = 3$

So, the slope $m$ of the tangent line at $x = 1$ is 3.

Step 1.4: Equation of the Tangent Line

Using the point-slope form $y - f(1) = m(x - 1)$, we get: $y - 9 = 3(x - 1)$ $y = 3x + 6$

Thus, the tangent line equation is: $L(x) = 3x + 6$

Task 2: Use the Tangent Line to Approximate $f(1.1)$

To approximate $f(1.1)$ using the tangent line, substitute $x = 1.1$ into $L(x)$: $L(1.1) = 3(1.1) + 6 = 3.3 + 6 = 9.3$

So, $L(1.1) \approx 9.3$.

Task 3: Compute the Actual Value of $f(1.1)$ and the Error

Step 3.1: Actual Value $f(1.1)$

Calculate $f(1.1)$: $f(1.1) = \sqrt[3]{243(1.1)^3 + 486}$ First, compute $243 \cdot (1.1)^3$: $243 \cdot 1.331 = 323.133$ Then add 486: $323.133 + 486 = 809.133$ Now, take the cube root: $f(1.1) \approx \sqrt[3]{809.133} \approx 9.2872$

Step 3.2: Compute the Error

The error is the absolute difference between $f(1.1)$ and $L(1.1)$: $| \text{error} | = | f(1.1) - L(1.1) | = | 9.2872 - 9.3 | \approx 0.0128$

Summary of Answers:

1. Equation of the tangent line: $L(x) = 3x + 6$
2. Approximation of $f(1.1)$ using the tangent line: $L(1.1) = 9.3$
3. Actual value of $f(1.1)$: $f(1.1) \approx 9.2872$
4. Error: $| \text{error} | \approx 0.0128$

Would you like further details or have any questions?

Additional Questions:

1. What is the significance of the tangent line in approximation?
2. How does the error help in understanding the accuracy of linear approximation?
3. Could the linear approximation be improved by using a higher-order derivative?
4. What is the geometric interpretation of the tangent line approximation?
5. How would the approximation change if evaluated at a different point?

Tip

In general, linear approximations work best for values of $x$ very close to the point where the tangent line is taken. The further you move from this point, the greater the error will likely be.