Problem 33 - Calculus Early Transcendentals

Question

Use the Mean-Value Theorem to prove that $$ \frac{x}{1+x^{2}}<\tan ^{-1} x0) $$

Step-by-Step Solution

Answer

The inequality holds: \( \frac{x}{1+x^2} < \tan^{-1}(x) < x \) for \( x > 0 \), using MVT and derivative properties.

Step 1: Understanding the Mean-Value Theorem (MVT)

The Mean-Value Theorem states that if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This theorem essentially states that the instantaneous rate of change (slope of the tangent) at some point is equal to the average rate of change over the interval.

Step 2: Setting Up Functions

Define \( f(x) = \tan^{-1}(x) \). We need to show \( \frac{x}{1+x^2} < \tan^{-1}(x) < x \) for \( x > 0 \). Note: \( \tan^{-1}(0) = 0 \), so both inequalities trivially hold for \( x = 0 \). Focus on \( x > 0 \).

Step 3: Prove the Right Inequality \( \tan^{-1}(x) < x \)

The derivative \( f'(x) = \frac{1}{1+x^2} \). For \( x > 0 \), we need \( f'(x) < 1 \), which simplifies to \( \frac{1}{1+x^2} < 1 \). Since \( 1+x^2 > 1 \) for \( x > 0 \), \( \frac{1}{1+x^2} < 1 \) holds. This implies \( \tan^{-1}(x) < x \) because the slope of the tangent line is less steep than that of the line \( y = x \).

Step 4: Prove the Left Inequality \( \frac{x}{1+x^2} < \tan^{-1}(x) \)

Now, rearrange terms to \( \tan^{-1}(x) \) by evaluating some interval using the Mean-Value Theorem. Consider the interval \([0, x]\). By MVT, there is some \( c \in (0, x) \) for which \( \frac{f(x) - f(0)}{x - 0} = f'(c) \). Since \( f'(x) = \frac{1}{1+x^2} \), \( f'(c) = \frac{1}{1+c^2} > \frac{1}{1+x^2} \). Thus, \( \frac{\tan^{-1}(x)}{x} > \frac{1}{1+x^2} \), so \( \tan^{-1}(x) > \frac{x}{1+x^2} \).

Step 5: Concluding the Proof

By proving both parts of the inequality separately, we established that \( \frac{x}{1+x^2} < \tan^{-1}(x) < x \) when \( x > 0 \), using properties from the Mean-Value Theorem and derivative analysis.

Key Concepts

Inequality Proofs

In mathematical analysis, proving inequalities often involves using established theorems and properties. The Mean-Value Theorem is particularly handy for these proofs. It helps compare different functions and their behaviors over an interval. To prove the inequality \( \frac{x}{1+x^2} < \tan^{-1}(x) < x \) for \( x > 0 \), we rely on the properties of the function \( \tan^{-1}(x) \).

• Firstly, understand the nature of the arctangent function. It is a strictly increasing function for \( x > 0 \).
• Then, consider its derivative \( f'(x) = \frac{1}{1+x^2} \), which describes the slope of the function.
• This helps in comparing it to the tangent function and bounding it between two simpler expressions.

By using these insights, you can establish the desired inequalities by aligning both derivatives and values of the functions precisely in the interval of interest.

Calculus Step-by-Step

To tackle the problem using the step-by-step approach in calculus, we begin with the Mean-Value Theorem. This theorem is powerful in analyzing the behavior of functions over a specific interval.

First, define the function of interest: \( f(x) = \tan^{-1}(x) \). Through calculus, we know this function will yield a derivative that dictates its slope: \( f'(x) = \frac{1}{1+x^2} \).

Using the Mean-Value Theorem for the interval \([0, x]\):

• Since \( f(x) \) is continuous on \([0, x]\) and differentiable on \((0, x)\), the conditions for the theorem are satisfied.
• The theorem tells us there is a \( c \) in \((0, x)\) where \( f'(c) = \frac{f(x) - f(0)}{x} \).

This analytical step finds a specific point where the instantaneous rate of change of the function (its derivative) equals the average rate over the interval.

Derivative Analysis

Derivative analysis is crucial to understanding the behavior and rate of change of functions. For the function \( \tan^{-1}(x) \), the derivative is given by \( f'(x) = \frac{1}{1+x^2} \).

Let's break down why knowing this derivative is important:

• The derivative \( \frac{1}{1+x^2} \) tells us how steep or flat the tangent line is at any point \( x \).
• For \( x > 0 \), it's always less than 1, indicating that the function \( \tan^{-1}(x) \) always increases at a rate slower than a linear function \( y = x \).
• This derivative helps contrast the function \( \tan^{-1}(x) \) both with \( x \) and \( \frac{x}{1+x^2} \).

By understanding how the derivative behaves, we ensure that the inequality \( \frac{x}{1+x^2} < \tan^{-1}(x) < x \) holds true. This is because it symbolizes how \( \tan^{-1}(x) \) smoothly fits between these two expressions across the positive domain.