Problem 34 - Thomas Calculus

Question

In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ g(x)=\sec x, \quad-\frac{\pi}{3} \leq x \leq \frac{\pi}{6} $$

Step-by-Step Solution

Answer

Absolute maximum is 2 at \(-\frac{\pi}{3}\), and absolute minimum is \(\frac{2}{\sqrt{3}}\) at \(\frac{\pi}{6}\).

Step 1: Understand the Function and Interval

We are given the function \( g(x) = \sec x \). The interval we are interested in is \( -\frac{\pi}{3} \leq x \leq \frac{\pi}{6} \). A secant function is defined as \( g(x) = \frac{1}{\cos x} \), so we need to ensure \( \cos x eq 0 \) within the interval.

Step 2: Identify Critical Points

To find where the maximum and minimum occur, we first need to look at the endpoints and critical points in the interval. Since \( g(x) \) is the secant of \( x \), and its derivative \( g'(x) = \sec x \tan x \), we set \( \sec x \tan x = 0 \) to find critical points. However, \( \sec x = 0 \) has no solutions, so only look for points where \( \tan x = 0 \), which doesn't occur because \( x = 0 \) isn't in the interval.

Step 3: Evaluate Endpoints of the Interval

Now calculate the function value at the endpoints:- At \( x = -\frac{\pi}{3} \): \( g(-\frac{\pi}{3}) = \sec(-\frac{\pi}{3}) = 2 \). - At \( x = \frac{\pi}{6} \): \( g(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \approx 1.1547 \).

Step 4: Determine Absolute Extrema

The absolute maximum value of \( g(x) \) is \( 2 \) at \( x = -\frac{\pi}{3} \), and the absolute minimum value is \( \frac{2}{\sqrt{3}} \approx 1.1547 \) at \( x = \frac{\pi}{6} \).

Step 5: Graph the Function

Graph the function \( g(x) = \sec x \) within the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\). Indicate the points \((-\frac{\pi}{3}, 2)\) and \((\frac{\pi}{6}, 1.1547)\) on the graph, as these are where the absolute maximum and minimum occur, respectively.

Key Concepts

Secant Function

In mathematics, the secant function is a trigonometric function, denoted as \( \sec x \). It is defined as the reciprocal of the cosine function, meaning \( \sec x = \frac{1}{\cos x} \). This function is undefined whenever the cosine of \( x \) is zero, so we need to be cautious about its domain.
The behavior of the secant function is closely tied to the behavior of the cosine function. For example:

• When \( \cos x \) is positive, \( \sec x \) will also be positive.
• When \( \cos x \) is negative, \( \sec x \) will be negative.
• The secant function experiences vertical asymptotes where \( \cos x = 0 \).

In general, the secant function is periodic with a period of \( 2\pi \). Understanding these properties can help you graph the function and find key features such as maximum and minimum values.

Derivative

To find where extreme values occur in a function, we often need to find its derivative. The derivative of the secant function \( g(x) = \sec x \) is \( g'(x) = \sec x \tan x \). This derivative tells us how the function is changing at any point.

• The derivative's value indicates the slope at a particular point on the function's graph.
• If \( g'(x) > 0 \), the function is increasing at that point.
• If \( g'(x) < 0 \), the function is decreasing.
• Critical points occur where \( g'(x) = 0 \) or is undefined.

In the given step solution, the derivative \( g'(x) = \sec x \tan x \) was set to zero to find critical points, but since this function has no solutions for \( \tan x = 0 \) within the specified interval, there are no additional critical points beyond the endpoints.

Critical Points

Critical points are places on the graph where

• the derivative of the function is zero (stationary points), or
• the derivative does not exist.

These points are crucial because they can potentially be locations of local maxima or minima. However, not all critical points will result in maximum or minimum values; some may correspond to inflection points where the curvature changes.
In our specific example with \( g(x) = \sec x \), when looking for critical points within the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\), none exist. As a result, we focus on the endpoints to evaluate the absolute extrema. This highlights that understanding the nature of critical points helps prioritize where to check for extreme functions.

Endpoints Evaluation

In many calculus problems, especially when dealing with closed intervals, it is essential to evaluate the function at the endpoints. This is where potential extreme (maximum or minimum) values might occur, especially if no critical points are found within the interval.

• To perform an endpoints evaluation, simply plug the endpoint values into the function.
• Compare these values to determine the highest and lowest values (absolute maximum and minimum).

In the example of the secant function, evaluating at the two endpoints, \(-\frac{\pi}{3}\) and \(\frac{\pi}{6}\), provided us with the necessary maximum and minimum values. At \( x = -\frac{\pi}{3} \), the secant function equals \( 2 \), which is the maximum value. At \( x = \frac{\pi}{6} \), it equals approximately \( 1.1547 \), the minimum value. Endpoint analysis is therefore a straightforward yet vital final step to finding absolute extremities in calculus problems.