We are given the function:

$f(x) = 1 + 2 \cos(\pi x) + \frac{1}{2}$

First, simplify the constant terms:

$f(x) = 1 + \frac{1}{2} + 2 \cos(\pi x) = \frac{3}{2} + 2 \cos(\pi x)$

Now, we analyze the term $2 \cos(\pi x)$. The cosine function $\cos(\pi x)$ always has a range between $-1$ and $1$, i.e.,

$-1 \leq \cos(\pi x) \leq 1$

Multiply this inequality by 2:

$-2 \leq 2 \cos(\pi x) \leq 2$

Now, add $\frac{3}{2}$ to the entire inequality:

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

Simplify both sides:

$-\frac{1}{2} \leq f(x) \leq \frac{7}{2}$

Thus, the range of $f(x)$ is:

$\left[ -\frac{1}{2}, \frac{7}{2} \right]$

Would you like further explanation or details on any of the steps?

Related questions:

1. What is the period of $\cos(\pi x)$?
2. How do the constants affect the range of trigonometric functions?
3. What is the maximum and minimum value of $f(x)$ for specific $x$?
4. How can we find the range of other transformed cosine functions?
5. Can the range change if we add more terms involving trigonometric functions?

Tip: When transforming trigonometric functions, always start by determining the range of the core function before applying any shifts or scaling.