Let's go through the questions one by one based on the image provided:

1. Given $f(x) = \frac{x+1}{x}$:

• (a) Find $f(10)$:
  Substitute $x = 10$ into $f(x)$: $f(10) = \frac{10+1}{10} = \frac{11}{10}.$

• (b) Find $\frac{f(3+h) - f(3)}{h}$:
  First, calculate $f(3+h)$ and $f(3)$:

  $f(3+h) = \frac{(3+h)+1}{3+h} = \frac{4+h}{3+h},$ $f(3) = \frac{3+1}{3} = \frac{4}{3}.$ Now substitute them into the expression: $\frac{f(3+h) - f(3)}{h} = \frac{\frac{4+h}{3+h} - \frac{4}{3}}{h}.$ This can be simplified further, but it's already the form of the difference quotient.

2. Determine the domain for each function:

• (a) $f(x) = \frac{x}{x^2 - 1}$:
  The denominator should not be zero, so: $x^2 - 1 \neq 0 \quad \Rightarrow \quad x \neq \pm1.$ Thus, the domain is $x \in (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

• (b) $g(x) = \frac{\sqrt{1 + x^2}}{2x + 3}$:
  For $g(x)$, the square root function requires the argument inside to be non-negative, and the denominator should not be zero: $2x + 3 \neq 0 \quad \Rightarrow \quad x \neq -\frac{3}{2}.$ The domain is $x \in \mathbb{R} \setminus \left\{-\frac{3}{2}\right\}$.

3. Determine whether the functions are odd, even, or neither:

• (a) $f(x) = \frac{3x}{x^2 + 1}$:
  Check if $f(-x) = -f(x)$ for odd or $f(-x) = f(x)$ for even: $f(-x) = \frac{-3x}{(-x)^2 + 1} = \frac{-3x}{x^2 + 1} = -f(x),$ so $f(x)$ is odd.

• (b) $g(x) = \frac{x^2 + 1}{|x| + x^4}$:
  This function is neither odd nor even because the presence of the absolute value and the powers of $x$ leads to asymmetry.

4. Given $f(x) = \frac{x - 1}{x}$ and $g(x) = x^2 + 1$:

• (a) $(f + g)(2)$:
  $f(2) = \frac{2-1}{2} = \frac{1}{2}, \quad g(2) = 2^2 + 1 = 5,$ $(f + g)(2) = f(2) + g(2) = \frac{1}{2} + 5 = \frac{11}{2}.$

• (b) $(f \cdot g)(2)$:
  $(f \cdot g)(2) = f(2) \cdot g(2) = \frac{1}{2} \cdot 5 = \frac{5}{2}.$

• (c) $(f \circ g)(2)$:
  First, calculate $g(2)$, then substitute it into $f(x)$: $g(2) = 5, \quad f(5) = \frac{5 - 1}{5} = \frac{4}{5}.$ So, $(f \circ g)(2) = \frac{4}{5}$.

• (d) $(g \circ f)(2)$:
  First, calculate $f(2)$, then substitute it into $g(x)$: $f(2) = \frac{1}{2}, \quad g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4}.$ So, $(g \circ f)(2) = \frac{5}{4}$.

• (e) $f^2(2) + g^2(2)$:
  This means $f(2)^2 + g(2)^2$: $f(2)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad g(2)^2 = 5^2 = 25,$ $f^2(2) + g^2(2) = \frac{1}{4} + 25 = \frac{101}{4}.$

Would you like to go into more detail or ask about any specific parts? Here are 5 related questions for further practice:

1. How would the answer to question 3(a) change if the function were $f(x) = \frac{x}{x^2 + 1}$?
2. How would you find the inverse of the function $f(x) = \frac{x+1}{x}$?
3. What is the significance of an odd function in terms of symmetry?
4. How would the domain of $f(x) = \frac{x}{x^2 - 4}$ change compared to question 2(a)?
5. How would you graph the function $g(x) = \frac{\sqrt{1+x^2}}{2x+3}$?

Tip: When calculating composition functions, always apply the inner function first!