Answer:
(a) L(x) = 5x + 1
(b) L(1.1) = 6.5
(c) error ≈ 0.00753
Explanation:
To address the question, we will follow a systematic approach to find the equation of the tangent line to the function at x = 1, use the tangent line to approximate the value of the function at x = 1.1, and then compute the actual value of the function at x = 1.1 to determine the error in the linear approximation.
Here is our given function:
![f(x) = \sqrt[3]{180x^3+36}](https://img.qammunity.org/2024/formulas/mathematics/college/pok67uf6uba1dwycaq9n2f0ouc92jkkwi9.png)
Note: At x = 1, f(1) = 6

Firstly, we calculate the derivative of the given function, which represents the slope of the tangent line at any point 'x'.
![\Longrightarrow (d)/(dx)\left[f(x) = \sqrt[3]{180x^3+36}\right]\\\\\\\\\Longrightarrow (d)/(dx)[f(x)] = (d)/(dx)\left[\sqrt[3]{180x^3+36}\right]\\\\\\\\\Longrightarrow f'(x)= (d)/(dx)\left[(180x^3+36)^{(1)/(3)}\right]\\\\\\\\\Longrightarrow f'(x)= (1)/(3) (180x^3+36)^{(1)/(3)-1} \cdot (d)/(dx)[180x^3+36]\\\\\\\\\Longrightarrow f'(x)= (1)/(3) (180x^3+36)^{-(2)/(3)} \cdot(540x^2)\\\\\\\\](https://img.qammunity.org/2024/formulas/mathematics/college/g2m9zqwxhrai5ttf14annkjm1knyqip4b9.png)

Plugging in x = 1 to find the slope:

With this slope, we find the equation of the tangent line at x = 1 by using the point-slope form, y - y₁ = m(x - x₁).
We have,
Plugging our values in:

Thus, our tangent line at x = 1 is L(x) = 5x + 1. Using our equation of the tangent line, plug in x = 1.1 to approximate the value of 'y':

To determine the actual value of 'y' plug in x = 1.1 into f(x) and approximate using a calculator:
![\Longrightarrow f(1.1) = \sqrt[3]{180(1.1)^3+36}\\\\\\\\\therefore f(1.1) \approx 6.50753 \text{ (rounded to 5 d.p.)}](https://img.qammunity.org/2024/formulas/mathematics/college/werf3zj3z3c71ydqldmiroalk3z09xoanq.png)
We can calculate the error as follows:

Thus, the error is approximately 0.00753.