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We are given the function:

\[
f(x) = \frac{x+1}{x}
\]

The tasks are:

**a.** Calculate \( f(10) \).  
**b.** Simplify \( \frac{f(3+h) - f(3)}{h} \).

### Solution:

#### a. \( f(10) \):

\[
f(x) = \frac{x+1}{x}
\]

Substituting \( x = 10 \):

\[
f(10) = \frac{10+1}{10} = \frac{11}{10}
\]

So, \( f(10) = \frac{11}{10} \).

#### b. Simplify \( \frac{f(3+h) - f(3)}{h} \):

We will first calculate \( f(3+h) \) and \( f(3) \), and then substitute into the expression.

1. **Find \( f(3+h) \):**

\[
f(3+h) = \frac{(3+h) + 1}{3+h} = \frac{4+h}{3+h}
\]

2. **Find \( f(3) \):**

\[
f(3) = \frac{3+1}{3} = \frac{4}{3}
\]

Now, substitute these into \( \frac{f(3+h) - f(3)}{h} \):

\[
\frac{f(3+h) - f(3)}{h} = \frac{\frac{4+h}{3+h} - \frac{4}{3}}{h}
\]

To simplify this, we need a common denominator for the two fractions:

\[
\frac{4+h}{3+h} - \frac{4}{3} = \frac{3(4+h) - 4(3+h)}{3(3+h)}
\]

Simplifying the numerator:

\[
3(4+h) - 4(3+h) = (12 + 3h) - (12 + 4h) = -h
\]

Thus, the expression becomes:

\[
\frac{-h}{3(3+h)} = \frac{-1}{3(3+h)}
\]

Now substitute this back into the original expression:

\[
\frac{\frac{-1}{3(3+h)}}{h} = \frac{-1}{h \cdot 3(3+h)} = \frac{-1}{3h(3+h)}
\]

Thus, the simplified form of \( \frac{f(3+h) - f(3)}{h} \) is:

\[
\frac{-1}{3h(3+h)}
\]

### Summary:
- \( f(10) = \frac{11}{10} \)
- \( \frac{f(3+h) - f(3)}{h} = \frac{-1}{3h(3+h)} \)

Let me know if you need further clarification or details!

---

Here are five related questions for further practice:
1. What is the derivative of \( f(x) = \frac{x+1}{x} \)?
2. Evaluate \( f(5) \) for the same function.
3. What is the limit of \( \frac{f(3+h) - f(3)}{h} \) as \( h \to 0 \)?
4. How would you simplify \( \frac{x^2 + x + 1}{x} \)?
5. Find \( f'(x) \) for \( f(x) = \frac{x^2 + 2}{x} \).

**Tip:** When simplifying rational functions, always look for common denominators to combine terms effectively!

Misalkan f(x) = (x+1)/x. Cari dan sederhanakan.
a. f(10)
b. f(3+h)-f(3)/h

Learn how to evaluate f(10) for the function f(x) = (x+1)/x and simplify the difference quotient (f(3+h) - f(3))/h. This problem covers function evaluation and simplifying expressions using algebraic techniques. Suitable for high school students.

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