The problem involves a function $f(x) = \frac{x^2 + 1}{2x}$. We are tasked to find:

1. $f(x) + f(3)$,
2. $f(x+3)$.

Step 1: Simplify $f(x)$

The given function is: $f(x) = \frac{x^2 + 1}{2x}.$

Step 2: Evaluate $f(3)$

Substitute $x = 3$ into $f(x)$: $f(3) = \frac{3^2 + 1}{2 \cdot 3} = \frac{9 + 1}{6} = \frac{10}{6} = \frac{5}{3}.$

Step 3: Find $f(x) + f(3)$

The sum is: $f(x) + f(3) = \frac{x^2 + 1}{2x} + \frac{5}{3}.$

To combine the terms, find a common denominator, which is $6x$: $f(x) + f(3) = \frac{3(x^2 + 1)}{6x} + \frac{10x}{6x} = \frac{3x^2 + 3 + 10x}{6x}.$ Thus: $f(x) + f(3) = \frac{3x^2 + 10x + 3}{6x}.$

Step 4: Find $f(x+3)$

Substitute $x+3$ into $f(x)$: $f(x+3) = \frac{(x+3)^2 + 1}{2(x+3)}.$

Expand $(x+3)^2$: $(x+3)^2 = x^2 + 6x + 9.$

So: $f(x+3) = \frac{x^2 + 6x + 9 + 1}{2(x+3)} = \frac{x^2 + 6x + 10}{2(x+3)}.$

Final Results

1. $f(x) + f(3) = \frac{3x^2 + 10x + 3}{6x}$,
2. $f(x+3) = \frac{x^2 + 6x + 10}{2(x+3)}$.

Let me know if you'd like me to explain any specific part or provide further details!

Here are 5 related questions:

1. What happens to $f(x)$ as $x \to \infty$?
2. How would $f(x)$ change if the numerator was $x^2 - 1$?
3. Can $f(x)$ be simplified further if $x \neq 0$?
4. What is the domain of $f(x)$?
5. How does $f(x)$ behave near $x = 0$?

Tip:

Always check the domain of a function before simplifying or substituting values. For this function, $x = 0$ is excluded.