https://file.aiquickdraw.com/mathbot/file/17273240879236i6hwgy6.jpg

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 \( f(2)^2 + g(2)^2 \):
  \[
  f(2)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad g(2)^2 = 5^2 = 25,
  \]
  \[
  f^2(2) + g^2(2) = \frac{1}{4} + 25 = \frac{101}{4}.
  \]

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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!

Misalkan f(x) = (x+1)/x. Cari dan sederhanakan. Tentukan daerah asal alami untuk setiap fungsi berikut. Tentukan fungsi-fungsi berikut fungsi ganjil, fungsi genap, atau tidak keduanya. Misalkan f(x) = x-1/x dan g(x) = x^2 + 1. Tentukan (f+g)(2), (f*g)(2), (fog)(2), (gof)(2), f^2(2) + g^2(2).

Dive into advanced function concepts including evaluation at specific points, determining the domain, identifying symmetry as odd or even, and performing function compositions. Suitable for high school students in Grades 11-12.

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