The problem asks for the value of $k$ when the function $f(x) = \frac{1}{2}(x + 3)^2$ is rewritten in standard form as:

$f(x) = \frac{1}{2}x^2 + kx + \frac{9}{2}$

To solve this, we need to expand the given expression and match it to the standard form.

Step 1: Expand $\frac{1}{2}(x + 3)^2$

We start by expanding $(x + 3)^2$:

$(x + 3)^2 = x^2 + 6x + 9$

Now multiply by $\frac{1}{2}$:

$f(x) = \frac{1}{2}(x^2 + 6x + 9)$

Distribute the $\frac{1}{2}$:

$f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2}$

Step 2: Compare with the standard form

Now, compare this result with the standard form:

$f(x) = \frac{1}{2}x^2 + kx + \frac{9}{2}$

We can see that the value of $k$ is $3$.

Thus, the value of $k$ is $\boxed{3}$.

Would you like more details on how to expand or any other concepts clarified? Here are a few related questions:

1. How do you expand a binomial square like $(x + 3)^2$?
2. What is the general method for converting a quadratic function into standard form?
3. How can you use completing the square to convert a quadratic into standard form?
4. What is the difference between vertex form and standard form of a quadratic function?
5. How would the value of $k$ change if the expression was $\frac{1}{3}(x + 3)^2$?

Tip: When expanding binomials, always remember to square both terms and apply the distributive property carefully.