To find the power series representation of the function $f(x) = \frac{1}{2} + 3x$ centered at $x_0 = -1$, we can express the function in the form:

$f(x) = \sum_{n=0}^{\infty} c_n (x - x_0)^n$

Here, $x_0 = -1$, so we want to write $f(x)$ as a power series in $(x + 1)$:

$f(x) = \sum_{n=0}^{\infty} c_n (x + 1)^n$

Given $f(x) = \frac{1}{2} + 3x$, we can write it as:

$f(x) = \frac{1}{2} + 3x = \frac{1}{2} + 3(x + 1 - 1) = \frac{1}{2} + 3[(x + 1) - 1]$

Simplifying, we get:

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

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

$f(x) = -\frac{5}{2} + 3(x + 1)$

Now, express this as a power series centered at $x_0 = -1$:

$f(x) = -\frac{5}{2} + 3(x + 1)$

So the power series is:

$f(x) = -\frac{5}{2} + 3(x + 1)$

Since this is a linear function, the series truncates after the first degree:

$f(x) = -\frac{5}{2} + 3(x + 1)$

Thus, the power series for $f(x) = \frac{1}{2} + 3x$ centered at $x_0 = -1$ is:

$f(x) = -\frac{5}{2} + 3(x + 1)$

This is the final answer, where the series has only two non-zero terms.

---

Would you like more details, or do you have any questions? Here are some related questions:

1. How do you find the power series for higher-degree polynomials?
2. What are the general steps to find the power series of any function centered at $x_0$?
3. Can the power series be used to approximate functions near $x_0$?
4. How does the radius of convergence affect the power series?
5. What happens if you center the series at a different $x_0$?

Tip: When working with power series, always consider the function's behavior around the center to ensure accuracy in approximation.