Problem 42 - Calculus Early Transcendentals

Question

If \(f\) is increasing on an interval \([0, b),\) then it follows from Definition 4.1 .1 that \(f(0)< f(x)\) for each \(x\) in the interval \((0, b) .\) Use this result in these exercises. Show that \(x< \tan x\) if \(0< x< \pi / 2,\) and confirm the inequality with a graphing utility. [Hint: Show that the function \(f(x)=\tan x-x \text { is increasing on }[0, \pi / 2) .]\)

Step-by-Step Solution

Answer

\(x < \tan x\) for \(0 < x < \pi/2\) because \(f(x) = \tan x - x\) is increasing on \([0, \pi/2)\). Graphically, \(y = \tan x\) is above \(y = x\) in this interval.

Step 1: Define the Function

Define the function we want to investigate: let \( f(x) = \tan x - x \). Our goal is to show that this function is increasing on the interval \( [0, \pi/2) \). This would imply that \( f(0) < f(x) \) for all \( x \) in \( (0, \pi/2) \).

Step 2: Differentiate the Function

Calculate the derivative \( f'(x) \) to determine whether the function is increasing in the given interval. We have \( f(x) = \tan x - x \), so the derivative is \( f'(x) = \sec^2 x - 1 \).

Step 3: Analyze the Derivative

Evaluate the derivative \( f'(x) = \sec^2 x - 1 \). Since \( \sec^2 x = 1/\cos^2 x \) and \( \cos x \leq 1 \) for \( 0 \leq x < \pi/2 \), \( \sec^2 x \geq 1 \), which implies \( \sec^2 x - 1 \geq 0 \). Thus, \( f'(x) \geq 0 \) on \( [0, \pi/2) \), showing that \( f(x) \) is increasing.

Step 4: Apply the Increasing Function Property

Since \( f(x) = \tan x - x \) is increasing, it follows that \( f(0) < f(x) \) for any \( x \in (0, \pi/2) \). Evaluating \( f(0) \), we have \( f(0) = \tan(0) - 0 = 0 \). Thus, \( 0 < \tan x - x \), which simplifies to \( x < \tan x \) for \( x \in (0, \pi/2) \).

Step 5: Confirm with Graphing Utility

Use a graphing utility to plot \( y = \tan x \) and \( y = x \) on the interval \( (0, \pi/2) \). The graph should confirm that the curve \( y = \tan x \) lies above \( y = x \), verifying the inequality \( x < \tan x \) in the specified interval.

Key Concepts

Increasing Functions

An increasing function is one where, as you move from left to right along the x-axis, the values of the function rise or stay constant. This means if you pick any two points, say \( a \) and \( b \), where \( a < b \), then \( f(a) \leq f(b) \). In mathematical terms, if a function is increasing on an interval \( [a, b] \), for each \( x_1 < x_2 \) in this interval, \( f(x_1) \leq f(x_2) \).

• If \( f(x_1) < f(x_2) \), the function is strictly increasing.
• If \( f(x_1) = f(x_2) \), it is constant on that subinterval.

Understanding whether a function is increasing is crucial when you want to establish inequalities like \( x < \tan x \). Since \( f(x) = \tan x - x \) is increasing in the interval \( [0, \pi/2) \), it supports the assertion that \( x < \tan x \) for any \( x \) in \( (0, \pi/2) \).
Analyzing and proving this property involves using derivatives, which gives us a solid mathematical ground to declare the behavior of functions without having to graph them initially.

Derivative Test

The derivative test is a powerful tool in calculus for determining the behavior of functions. It involves computing the derivative of a function, \( f'(x) \), and analyzing its sign across an interval.
If \( f'(x) > 0 \) on an interval, the function is increasing on that interval.
We've seen that for \( f(x) = \tan x - x \), the derivative is \( f'(x) = \sec^2 x - 1 \).

• Where \( \sec^2 x - 1 \geq 0 \), \( f(x) \) is increasing.
• \( \sec^2 x = 1/\cos^2 x \), and since \( \cos x \) does not equal zero in \( (0, \pi/2) \), \( \/\sec^2 x \) remains greater than 1.

For this interval, the derivative is always positive, confirming that \( f(x) = \tan x - x \) increases from \( 0 \) to \( \pi/2 \).
The derivative test provides a straightforward mathematical approach to interpreting the behavior of functions in given intervals without needing to visualize on a graph immediately.

Trigonometric Functions

Trigonometric functions like \( \tan x \) and \( \sec x \) are essential in calculus due to their periodic nature and diverse applications. Understanding the behavior of \( \tan x \) is crucial in exercises involving inequalities like \( x < \tan x \).
For \( x \in (0, \pi/2) \):

• \( \tan x \) is undefined at \( \pi/2 \) but increases continuously and steeply as \( x \) approaches this limit.
• \( \tan x \) becomes much larger than \( x \) particularly as x nears \(rac{\pi}{2}\)

Graphically, \( \tan x \) starts at zero and increases to infinity, illustrating why the inequality \( x < \tan x \) holds true for \( x \) in \( (0, \pi/2) \).
The derivative of \( \tan x \), known as \( \sec^2 x \), provides the rate of change, helping validate the increasing nature of \( \tan x - x \) analytically.

Graphing Utilities

Graphing utilities, like graphing calculators or software such as Desmos or GeoGebra, play an essential role in visualizing mathematical concepts. They are particularly useful in verifying inequalities visually, like \( x < \tan x \) for \( x \in (0, \pi/2) \).
By plotting functions like \( y = \tan x \) and \( y = x \) over an interval:

• You can visually check the relationship between the two functions.
• The graph will distinctly show \( y = \tan x \) lying above \( y = x \), confirming the inequality.

These visual tools also allow students to experiment with different intervals and conditions, enhancing their comprehension of calculus concepts. While analytical solutions are robust, graphing utilities provide intuitive insights, making it easier to understand function behavior at a glance.