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### Part (a): Prove that \( f(x) = x - \ln{x} \) is increasing for \( x > 1 \).

To prove that the function \( f(x) = x - \ln{x} \) is increasing for \( x > 1 \), we need to check whether the derivative \( f'(x) \) is positive for \( x > 1 \).

#### Step 1: Differentiate \( f(x) \)
\[
f'(x) = \frac{d}{dx}\left(x - \ln{x}\right) = 1 - \frac{1}{x}
\]

#### Step 2: Analyze the sign of \( f'(x) \)
\[
f'(x) = 1 - \frac{1}{x}
\]
For \( x > 1 \):
- \( \frac{1}{x} \) is a positive number and less than 1.
- Therefore, \( f'(x) = 1 - \frac{1}{x} \) will be positive for all \( x > 1 \) because \( 1 > \frac{1}{x} \) for \( x > 1 \).

Thus, \( f(x) = x - \ln{x} \) is an increasing function for \( x > 1 \) because its derivative is positive in this interval.

---

### Part (b): Using part (a), show that \( \ln{x} < x \) for \( x > 1 \).

Since we have already shown that \( f(x) = x - \ln{x} \) is increasing for \( x > 1 \), this implies that \( f(x) > f(1) \) for \( x > 1 \).

#### Step 1: Compute \( f(1) \)
\[
f(1) = 1 - \ln{1} = 1 - 0 = 1
\]

#### Step 2: Use the fact that \( f(x) \) is increasing
For \( x > 1 \), we know that \( f(x) > f(1) \), which means:
\[
x - \ln{x} > 1
\]
This simplifies to:
\[
x > \ln{x} \quad \text{for} \quad x > 1
\]

Thus, \( \ln{x} < x \) for \( x > 1 \), as required.

---

Would you like more details on any of these steps, or do you have any questions?

Here are some related questions to explore:
1. Can you find the second derivative of \( f(x) = x - \ln{x} \) and interpret its concavity?
2. How does the behavior of \( f(x) \) for \( x \leq 1 \) compare to that for \( x > 1 \)?
3. Can you explore the inequality \( \ln{x} < x - 1 \) for \( x > 1 \)?
4. How would the graph of \( f(x) = x - \ln{x} \) look for various values of \( x \)?
5. Can you prove the inequality \( x > \ln{x} \) for \( x > 0 \) without using derivatives?

**Tip**: When determining if a function is increasing or decreasing, analyzing the sign of its derivative is a useful approach.

a. Prove that f(x) = x - ln x is increasing for x > 1.
b. Using part (a), show that ln x < x for x > 1.

Explore the step-by-step solution proving that the function f(x) = x - ln(x) is increasing for x > 1, and how this result implies the inequality ln(x) < x. The solution covers key calculus concepts like differentiation and increasing function theorems, making it ideal for high school students in Grades 11-12.

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