Problem 34 - Thomas Calculus in SI Units

Question

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

The absolute maximum is 2 at \( x = -\frac{\pi}{3} \), and the absolute minimum is 1 at \( x = 0 \).

Step 1: Understand the Function and Domain

The function given is \( g(x) = \sec x \), and the domain is the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6} \). The function \( \sec x \) is defined as \( \frac{1}{\cos x} \). It's crucial to consider the behavior of \( \cos x \), especially around the interval limits, as \( \sec x \) will be undefined where \( \cos x = 0 \).

Step 2: Identify Critical Points

To find critical points within the interval, we need the derivative of \( g(x) = \sec x \). The derivative is \( g'(x) = \sec x \tan x \). Critical points occur where this derivative is zero or undefined. \( \sec x \tan x = 0 \) when \( \tan x = 0 \); however, within our interval, \( x = 0 \) is the only critical point.

Step 3: Evaluate the Function at Critical Points and Endpoints

Calculate \( g(x) = \sec x \) at the critical point and endpoints of the interval:- At \( x = -\frac{\pi}{3} \): \( g\left(-\frac{\pi}{3}\right) = \sec\left(-\frac{\pi}{3}\right) = 2 \).- At \( x = 0 \): \( g(0) = \sec 0 = 1 \).- At \( x = \frac{\pi}{6} \): \( g\left(\frac{\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \).

Step 4: Determine Absolute Extrema

From the evaluations, compare the values:- \( g\left(-\frac{\pi}{3}\right) = 2 \) is the largest value.- \( g(0) = 1 \) is the smallest value.- \( g\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \) is between these values.Thus, the absolute maximum is at \( x = -\frac{\pi}{3} \) and the absolute minimum is at \( x = 0 \).

Step 5: Graph the Function and Identify Points

Plot \( g(x) = \sec x \) over the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6} \). Mark the points where the extrema occur:- Absolute maximum: \((-\frac{\pi}{3}, 2)\).- Absolute minimum: \((0, 1)\).Ensure graphing takes into account the vertical asymptote beyond our interval.

Key Concepts

Absolute Extrema

In calculus, absolute extrema refer to the highest (maximum) and lowest (minimum) values that a function achieves over a specified interval. In the context of a function like \( g(x) = \sec x \), identified on the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\), absolute extrema involve finding these extreme values within the given domain.

To determine the absolute extrema, we evaluate the function at critical points and endpoints. This involves comparing these values to identify the absolute maximum and minimum:

• The Absolute Maximum occurs at the highest value, \( g\left(-\frac{\pi}{3}\right) = 2 \).
• The Absolute Minimum is at the lowest value, \( g(0) = 1 \).

Identifying these points helps in understanding the function's behavior across the interval.

Secant Function

The secant function, denoted as \( \sec x \), is the reciprocal of the cosine function: \( \sec x = \frac{1}{\cos x} \). This means it takes values based on the behavior of the cosine function. Whenever \( \cos x \) approaches zero, \( \sec x \) becomes very large, leading to vertical asymptotes in the graph.

In our exercise, we examine \( g(x) = \sec x \) over a specific interval. It's vital to keep in mind that the secant function is undefined at points where \( \cos x = 0 \). However, within the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\), \( \cos x \) does not reach zero, ensuring that \( \sec x \) remains defined.

Derivative

A derivative provides crucial insight into the rate of change of a function. For \( g(x) = \sec x \), the derivative \( g'(x) = \sec x \tan x \) helps us identify the critical points where the function's slope is zero or undefined.

In this exercise, the criticality is determined by setting the derivative to zero:

• We solve \( \sec x \tan x = 0 \), resulting in \( \tan x = 0 \).
• The only critical point within \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\) is \( x = 0 \).

This critical point, along with endpoints, is essential for finding absolute extrema.

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. They indicate potential locations where a function may experience relative maxima or minima.

For \( g(x) = \sec x \):

• We found the derivative, \( g'(x) = \sec x \tan x \).
• By setting it to zero, \( \tan x = 0 \) leads us to critical points.

In the interval \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\), the critical point is \( x = 0 \).
This analysis helps us decide the points where absolute extrema might occur and verifies using endpoint evaluations.