The given function is $f(x) = \tan(\pi x)$.

To find the vertical asymptotes of the function:

1. Recall the behavior of the tangent function: The tangent function $\tan(\theta)$ has vertical asymptotes wherever $\cos(\theta) = 0$, which corresponds to $\theta = \frac{\pi}{2} + n\pi$, where $n$ is any integer.

2. Apply this to the function $\tan(\pi x)$: We need to find where $\tan(\pi x)$ becomes undefined, which happens when $\cos(\pi x) = 0$.

3. Solve for $x$: $\cos(\pi x) = 0 \quad \text{when} \quad \pi x = \frac{\pi}{2} + n\pi.$ Divide by $\pi$: $x = \frac{1}{2} + n.$ Hence, the vertical asymptotes occur at $x = \frac{1}{2} + n$, where $n$ is any integer.

Conclusion:

The vertical asymptotes of the function $f(x) = \tan(\pi x)$ occur at $x = \frac{1}{2} + n$, where $n$ is an integer.

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Related Questions:

1. What are the horizontal asymptotes of the tangent function?
2. How does the graph of $\tan(\pi x)$ differ from $\tan(x)$?
3. What are the zeros of $f(x) = \tan(\pi x)$?
4. How do you determine the period of the function $\tan(\pi x)$?
5. Can you explain how to find the vertical asymptotes of $\sec(\pi x)$?

Tip:

For any trigonometric function like tangent, vertical asymptotes occur where the denominator of the associated sine and cosine function becomes zero.