Let's go through the solution step-by-step based on the provided information.

Step-by-Step Solution

1. Problem Overview: We are given the function $f(x) = \tan(\pi x)$ and need to estimate the value of $\tan(0.34 \pi)$ using the linear approximation at a nearby point $x = \frac{1}{3}$.

2. Step A: Compute the Derivative of $f(x) = \tan(\pi x)$ at $x = \frac{1}{3}$

   To find $f'(x)$, we differentiate $f(x) = \tan(\pi x)$: $f'(x) = \frac{d}{dx} \left( \tan(\pi x) \right) = \pi \sec^2(\pi x)$

   Now, substitute $x = \frac{1}{3}$: $f'\left(\frac{1}{3}\right) = \pi \sec^2\left(\frac{\pi}{3}\right)$

   We know that $\sec\left(\frac{\pi}{3}\right) = 2$, so: $f'\left(\frac{1}{3}\right) = \pi \cdot 2^2 = 4\pi$

   Thus, the answer for part (a) is: $f'\left(\frac{1}{3}\right) = 4\pi$

3. Step B: Write Down the Linearization $L(x)$ of $\tan(\pi x)$ at $x = \frac{1}{3}$

   The linearization $L(x)$ of $f(x)$ at $x = \frac{1}{3}$ is given by: $L(x) = f\left(\frac{1}{3}\right) + f'\left(\frac{1}{3}\right) \cdot \left(x - \frac{1}{3}\right)$

   We need to find $f\left(\frac{1}{3}\right)$: $f\left(\frac{1}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$

   Now, substitute $f\left(\frac{1}{3}\right) = \sqrt{3}$ and $f'\left(\frac{1}{3}\right) = 4\pi$ into the linearization formula: $L(x) = \sqrt{3} + 4\pi \left(x - \frac{1}{3}\right)$

   Therefore, the answer for part (b) is: $L(x) = \sqrt{3} + 4\pi \left(x - \frac{1}{3}\right)$

4. Step C: Estimate $\tan(0.34 \pi)$ Using Linear Approximation

   Since $0.34$ is close to $\frac{1}{3} \approx 0.3333$, we can use $L(x)$ to estimate $\tan(0.34 \pi)$ by setting $x = 0.34$: $L(0.34) = \sqrt{3} + 4\pi \left(0.34 - \frac{1}{3}\right)$

   Calculate $0.34 - \frac{1}{3}$: $0.34 - \frac{1}{3} = 0.34 - 0.3333 \approx 0.0067$

   Now, substitute back: $L(0.34) \approx \sqrt{3} + 4\pi \cdot 0.0067$

   Since $\pi \approx 3.14159$: $4\pi \cdot 0.0067 \approx 4 \cdot 3.14159 \cdot 0.0067 \approx 0.084$

   Thus: $L(0.34) \approx \sqrt{3} + 0.084$

   Using $\sqrt{3} \approx 1.73205$: $L(0.34) \approx 1.73205 + 0.084 = 1.81605$

   Rounded to six decimal places, the estimate is: $\tan(0.34 \pi) \approx 1.816050$

Let me know if you'd like further details on any of these steps, or if you have any questions!

Here are some related questions you might explore:

1. How does the linear approximation work for values farther from $x = 1/3$?
2. Why do we use the derivative in the linearization process?
3. How does the error in approximation change as we move farther from $x = 1/3$?
4. Can we improve this approximation by using higher-order terms (Taylor series)?
5. How would the approximation change if we estimated $\tan(0.35\pi)$ instead of $\tan(0.34\pi)$?

Tip: When estimating trigonometric values, using nearby known values and linearization can be a powerful tool for quick approximations.