Problem 72 - Thomas Calculus

Question

$$ \begin{array}{l}{\text { a. Prove that } f(x)=x-\text { ln } x \text { is increasing for } x >1} \\ {\text { b. Using part (a), show that } \ln x < x \text { if } x >1}\end{array} $$

Step-by-Step Solution

Answer

\( f(x) = x - \ln x \) is increasing for \( x > 1 \), proving that \( \ln x < x \) for \( x > 1 \).

Step 1: Determining the Derivative of f(x)

To determine where the function \( f(x) = x - \ln x \) is increasing, we first need to find its derivative. The derivative of \( f(x) \) is computed as follows: \( f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\ln x) = 1 - \frac{1}{x} \).

Step 2: Analyzing the Sign of the Derivative

The next step is to analyze \( f'(x) = 1 - \frac{1}{x} \). We want to find where this derivative is greater than zero since a function is increasing where its derivative is positive. Set \( 1 - \frac{1}{x} > 0 \). Solving this inequality, we get \( \frac{1}{x} < 1 \), or equivalently, \( x > 1 \). Therefore, \( f(x) \) is increasing for \( x > 1 \).

Step 3: Using the Increasing Property to Show \( \ln x 1 \)

Since we have shown that \( f(x) = x - \ln x \) is increasing for \( x > 1 \), it implies that \( f(x) > f(1) \) for \( x > 1 \). Calculate \( f(1) = 1 - \ln 1 = 1 - 0 = 1 \). Hence, \( x - \ln x > 1 \) for \( x > 1 \), meaning \( x > \ln x + 1 \). Thus, \( \ln x < x \) for \( x > 1 \).

Key Concepts

Understanding Increasing Functions

An increasing function is one where, as the input (or x-value) increases, the output (or y-value) also increases. To determine this, we often look at the derivative of a function. The derivative, denoted as \( f'(x) \), gives us a way to understand how a function is behaving at any point.

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.
• If \( f'(x) = 0 \), it might be a point of inflection or a local minimum/maximum.

For the function \( f(x) = x - \ln x \), its derivative \( f'(x) = 1 - \frac{1}{x} \) is crucial to determine where the function increases. By setting \( f'(x) > 0 \), we solve the inequality to find \( x > 1 \). This means the function \( f(x) = x - \ln x \) is increasing for all \( x > 1 \). Understanding this property helps us in various proofs and applications within calculus.

Exploring Inequality in Calculus

Inequalities are a fundamental part of calculus and mathematics in general. They help us understand relationships between different quantities. In calculus, we use inequalities to determine intervals where functions grow, shrink, or behave in particular ways.
When analyzing the function \( f(x) = x - \ln x \), we found that it increases when \( x > 1 \). This increasing behavior leads directly to an important inequality: \( \ln x < x \) for \( x > 1 \). How is this derived?
Given that \( f(x) \) increases beyond \( x = 1 \) and knowing it holds that \( f(x) > f(1) \), with \( f(1) = 1 \), the inequality translates to \( x - \ln x > 1 \), thus \( \ln x < x \). Such inequalities are powerful, as they provide bounds and relationships that are key to problem-solving in mathematics.

The Role of Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a special logarithmic function that uses the base \( e \) (approximately 2.718). It is defined for positive real numbers and has unique properties that make it essential in calculus.

• The derivative of \( \ln x \) is \( \frac{1}{x} \), which means it contributes a shrinking term to the derivative of the combined function \( f(x) = x - \ln x \).
• It grows slower as \( x \) increases compared to linear functions \( y = x \), which is why across certain domains, it will always be less than \( x \), as is seen in \( \ln x < x \) for \( x > 1 \).

The natural logarithm is important in many calculus problems, especially those involving exponential growth and decay, optimization problems, and in proofs such as this one. Its relationship with linear functions can also be compared graphically or algebraically to understand its behavior fully.

Purpose of Calculus Proofs

Calculus proofs are essential for the validation and understanding of mathematical concepts. They often rely on derivatives, inequalities, limits, and other fundamental concepts.
In this exercise, we used such a proof to show that \( \ln x < x \) for \( x > 1 \). Here's a breakdown:

• Start by analyzing the function \( f(x) = x - \ln x \). Determine where it has its increasing behaviour by using its derivative.
• Calculate the critical points and evaluate them to infer how the function behaves beyond these points, specifically beyond \( x = 1 \).
• Rely on the property of an increasing function to deduce valid inequalities.

Proofs like these not only cement the understanding of individual functions but also cultivate logical thinking and problem-solving skills. They ensure that we have a structured approach to tackling unfamiliar problems in calculus.