Problem 57 - Essential Calculus Early Transcendentals

Question

Show that \(\tan x>x\) for \(0< x<\pi / 2 .[\)Hint : Show that \(f(x)=\tan x-x\) is increasing on \((0, \pi / 2) . ]\)

Step-by-Step Solution

Answer

\( \tan x > x \) for \( 0 < x < \frac{\pi}{2} \) because \( \tan x - x \) is increasing and positive.

Step 1: Define the Function

Let \( f(x) = \tan x - x \). Our goal is to show that \( f(x) > 0 \) for \( 0 < x < \frac{\pi}{2} \).

Step 2: Find the Derivative

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

Step 3: Simplify the Derivative

Since \( \sec^2 x = 1 + \tan^2 x \), we can simplify \( f'(x) = \sec^2 x - 1 = \tan^2 x \).

Step 4: Analyze the Sign of the Derivative

Notice that \( \tan^2 x \) is positive for \( 0 < x < \frac{\pi}{2} \). Thus, \( f'(x) > 0 \) on \( (0, \frac{\pi}{2}) \).

Step 5: Conclude the Increasing Nature of f(x)

Since \( f'(x) > 0 \) on \( (0, \frac{\pi}{2}) \), the function \( f(x) = \tan x - x \) is strictly increasing on this interval.

Step 6: Initial Value Analysis

Evaluate \( f(x) \) at the lower bound: \( f(0) = \tan 0 - 0 = 0 \). Since \( f(x) \) is strictly increasing, \( f(x) > 0 \) for all \( x \) in \( 0 < x < \frac{\pi}{2} \). Thus, \( \tan x > x \).

Key Concepts

Derivative Analysis

To understand calculus inequalities, it's crucial to perform derivative analysis. Let's apply this to the function \( f(x) = \tan x - x \). The main aim is to determine whether this function is increasing on the interval \((0, \frac{\pi}{2})\). To do this, we start by finding the derivative of the function, which gives us a tool to analyze its behavior.

• The derivative of \( \tan x \) is \( \sec^2 x \).
• The derivative of \( x \) is simply 1.
• Thus, the derivative of \( f(x) \) is \( f'(x) = \sec^2 x - 1 \).

Simplifying \( f'(x) \) further, remember that \( \sec^2 x = 1 + \tan^2 x \). Therefore, \( f'(x) = \sec^2 x - 1 = \tan^2 x \). This simplification tells us that the derivative is expressed purely in terms of \( \tan x \), which is positive for all \( x \) in the given interval.

Increasing Functions

An increasing function is one where as \( x \) increases, \( f(x) \) also increases. We determine whether a function is strictly increasing by checking if its derivative is positive over the interval of interest.
In our case, we found the derivative \( f'(x) = \tan^2 x \). We know that \( \tan^2 x > 0 \) for \( 0 < x < \frac{\pi}{2} \).

• This positivity means that the slope of the tangent at every point in the interval is upward.
• This is a clear indicator that \( f(x) \) is strictly increasing on \((0, \frac{\pi}{2})\).

Since \( f(x) = \tan x - x \) starts at \( f(0) = 0 \), and because \( f(x) \) is increasing, it follows that \( f(x) > 0 \) for all \( x \) in \( 0 < x < \frac{\pi}{2} \). Therefore, this confirms the inequality \( \tan x > x \) within this interval.

Trigonometric Inequalities

Trigonometric inequalities often involve comparing a trigonometric function with another function or constant over an interval. Here, we analyze \( \tan x > x \) for \( 0 < x < \frac{\pi}{2} \). The strategy involves defining a function such as \( f(x) = \tan x - x \), and then proving it remains positive.
Firstly, we used derivative analysis to determine that \( f(x) \) is strictly increasing over the interval. This approach helps by verifying that starting from an initial value, the function only goes upwards.

• Since \( f(0) = 0 \) and \( f(x) \) is increasing, it ensures \( f(x) > 0 \) throughout the interval \( 0 < x < \frac{\pi}{2} \).
• Thus, it supports the conclusion that \( \tan x \), being larger, rises faster than \( x \) in this specific range.

This entire process illustrates how the understanding of calculus and trigonometric properties help solve inequalities, broadening our mathematical insight and problem-solving skills.