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.