Let's analyze the problem step by step.

Given Function:

$f(x) = \frac{x}{2x+3}$

Part (a): Find the first Taylor polynomial $P_1(x)$ of $f(x)$ at $x = 0$.

The Taylor series expansion of a function $f(x)$ around $x = a$ is given by: $f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots$

Since we are looking for the first Taylor polynomial $P_1(x)$, we only need the first two terms of the series: $P_1(x) = f(0) + f'(0)x$

1. Find $f(0)$: $f(0) = \frac{0}{2(0) + 3} = 0$

2. Find $f'(x)$: Use the quotient rule: $f'(x) = \frac{(2x + 3)(1) - x(2)}{(2x + 3)^2} = \frac{2x + 3 - 2x}{(2x + 3)^2} = \frac{3}{(2x + 3)^2}$ Now evaluate $f'(0)$: $f'(0) = \frac{3}{(2(0) + 3)^2} = \frac{3}{9} = \frac{1}{3}$

Thus, the first Taylor polynomial is: $P_1(x) = 0 + \frac{1}{3}x = \frac{1}{3}x$

Part (b): Use the Taylor polynomial to approximate $f(1.05)$.

Now, approximate $f(1.05)$ using $P_1(x)$: $P_1(1.05) = \frac{1}{3} \times 1.05 = \frac{1.05}{3} \approx 0.35$

Part (c): Calculate the error in the Taylor series approximation, $|f(1.05) - P_1(1.05)|$, to 4 decimal places.

1. Calculate $f(1.05)$: $f(1.05) = \frac{1.05}{2(1.05) + 3} = \frac{1.05}{2.1 + 3} = \frac{1.05}{5.1} \approx 0.2059$

2. Calculate the error: $\text{Error} = |f(1.05) - P_1(1.05)| = |0.2059 - 0.35| = 0.1441$

So, the error is approximately 0.1441.

Would you like any further details or explanations on these calculations?

Here are five related questions:

1. What is the second-degree Taylor polynomial $P_2(x)$ for $f(x)$ at $x = 0$?
2. How does the error change if you use the second-degree polynomial instead of the first-degree polynomial?
3. What is the radius of convergence for the Taylor series of $f(x)$ around $x = 0$?
4. How would the Taylor polynomial change if you expanded around $x = 1$ instead of $x = 0$?
5. How does the Taylor series help in approximating functions that are difficult to evaluate directly?

Tip: Remember that the error in a Taylor series approximation depends on the number of terms used. The more terms you include, the smaller the error, especially near the point of expansion.