Problem 62 - Calculus with Concepts in Calculus

Question

Show that \(\tan x>x\) for all \(x\) in \((0, \pi / 2)\).

Step-by-Step Solution

Answer

\( \tan(x) > x \) for all \( x \) in \((0, \pi/2)\).

Step 1: Understand the Problem

We need to prove that the tangent function is greater than the line represented by the function f(x) = x for all x within the interval (0, π/2). Mathematically, this is \( \tan(x) > x \) for all x in \( (0, \pi / 2)\).

Step 2: Consider the Function f(x) = tan(x) - x

Define a new function \( f(x) = \tan(x) - x \). We need to demonstrate that f(x) > 0 for all x in (0, π/2).

Step 3: Find the Derivative of f(x)

Compute the derivative of f(x): \( f'(x) = \frac{d}{dx}(\tan(x)) - \frac{d}{dx}(x) = \sec^2(x) - 1 \).

Step 4: Analyze the Derivative

The derivative \( f'(x) = \sec^2(x) - 1 \) is positive if \( \sec^2(x) > 1 \). Since \( \sec(x) = \frac{1}{\cos(x)} \), \( \sec^2(x) = \frac{1}{\cos^2(x)} \), and given that \( 0 < x < \frac{\pi}{2} \), \( \cos(x) < 1 \). Therefore, \( \sec^2(x) > 1 \), thus \( f'(x) > 0 \) for all x in (0, π/2).

Step 5: Determine the Behavior of f(x)

Since \( f(x) \) is increasing and \( f(x) > 0 \) on \( (0, \pi/2) \) would imply \( \tan(x) > x \), evaluate in the limit as x approaches 0 from the right, \( \lim_{x \to 0^+} \tan(x) = 0 = \lim_{x \to 0^+} x \.\) Hence, at \( x \to 0^+ \), the behavior confirms that the slope is positive from this point on.

Step 6: Conclude the Proof

Since \( f'(x) > 0 \) on \( (0, \pi/2) \), \( f(x) = \tan(x) - x \) is strictly increasing on \( (0, \pi/2) \). Given that \( f(x) \) starts at 0 and increases, \( f(x) > 0 \), corroborating that \( \tan(x) > x \) for all x in \( (0, \pi/2)\).

Key Concepts

Derivative Analysis

Analyzing the derivative of a function is key to understanding its behavior. Here, we work with the function \( f(x) = \tan(x) - x \) and find its derivative to determine whether \( \tan(x) \) is greater than \( x \). The derivative of \( f(x) \) is \( f'(x) = \sec^2(x) - 1 \). This analysis helps us identify whether \( f(x) \) is increasing or decreasing within a specified interval, such as \( (0, \pi/2) \).

When we say \( f'(x) = \sec^2(x) - 1 \), it is clear that for \( \sec^2(x) > 1 \), \( f'(x) \) is positive. This positivity implies that \( f(x) \) is increasing. Thus, by studying \( f'(x) \), we gain insight into the slope of the graph of \( f(x) \), which indicates whether the original function \( \tan(x) \) minus \( x \) is greater than zero. Typically, identifying where a function increases or decreases assists in determining properties like maximums, minimums, or inequalities, which is vital to solve the problem.

• A positive derivative indicates an increasing function.
• A zero or negative derivative indicates constant or decreasing function respectively.

Increasing Functions

An increasing function is one where, as the input (usually \( x \)) increases, the output (usually \( f(x) \)) also increases. For the function \( f(x) = \tan(x) - x \), we have already determined that the derivative \( f'(x) = \sec^2(x) - 1 \) is positive for all \( x \) in \( (0, \pi/2) \).

This tells us that \( f(x) \) is increasing on this interval, meaning if you take any two points, say \( a \) and \( b \) with \( a < b \), then \( f(a) < f(b) \). This is crucial because it implies that the initial point is lower than subsequent points, confirming that the function always increases within the interval. Thus, the inequality \( \tan(x) > x \) holds, as the function starts from zero and increases continuously.

• For an increasing function, if \( x_1 < x_2 \), then \( f(x_1) < f(x_2) \).
• This property helps prove inequalities by showing that one function consistently exceeds another over a range.

Trigonometric Functions

Trigonometric functions like \( \tan(x) \) play a pivotal role in calculus and many mathematical proofs. These functions often relate to angles and the ratios within triangles, but their applications extend far beyond.

The tangent function, \( \tan(x) \), is defined as the ratio of the sine and cosine functions: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). This description means it is undefined where \( \cos(x) = 0 \), and in the range \( (0, \pi/2) \), \( \tan(x) \) is always positive and increases from 0 to infinity as \( x \) approaches \( \pi/2 \).

• \( \tan(x) \) is positive and increasing in the interval \( (0, \pi/2) \).
• Understanding the behavior of \( \tan(x) \) allows us to analyze equations and inequalities involving trigonometric functions.
• When combined with derivative analysis, we can determine the conditions under which trigonometric functions exceed certain values.