Problem 28 - Essential Calculus

Question

Locate the absolute extrema of the function on the closed interval. $$ g(x)=\sec x,\left[-\frac{\pi}{6}, \frac{\pi}{3}\right] $$

Step-by-Step Solution

Answer

The absolute minimum of the function on the interval is \(2 / \sqrt{3}\) at \(x = - \pi / 6\) and the absolute maximum is \(2\) at \(x = \pi / 3\).

Step 1: Differentiate the function

First, compute the derivative of \( g(x) = \sec x \). The derivative of \(\sec x\) is \( g'(x) = \sec x \tan x\).

Step 2: Find critical points

Next, set the derivative equal to zero to find any critical points. However, \(\sec x \tan x = 0\) has no solutions since for all x, \(\sec x \neq 0 \) and \(\tan x \neq 0\). So there are no critical points in the given interval.

Step 3: Evaluate the function at the end points

Since no critical points exist, we need to evaluate the function at the ends of the given interval, \(x= -\frac{\pi}{6}\) and \(x= \frac{\pi}{3}\): g(-\frac{\pi}{6}) and g(\frac{\pi}{3}). Thus, \( g(-\pi / 6) = 2 / \sqrt{3} \) and \( g(\pi / 3) = 2 \).

Step 4: Determine the absolute extrema

Compare the y-values at the end points and any critical values within the interval (if exist) to determine the maximum and the minimum. In this case, \( g(-\pi / 6) = 2 / \sqrt{3} < g(\pi / 3) = 2 \). Hence, the absolute minimum of g on the interval \([- \pi / 6, \pi / 3]\) is \(2 / \sqrt{3}\) which occurs at \(x = - \pi / 6\) and the absolute maximum is \(2\) which occurs at \(x = \pi / 3\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate of change of a function. This rate of change is represented by the derivative of the function. When differentiating a function, you essentially determine how the function's output changes as the input changes. Here, for the function

• \( g(x) = \sec x \), the derivative is \( g'(x) = \sec x \tan x \).

When differentiating trigonometric functions like secant, it is important to understand the derivatives of standard trigonometric functions:

• \( \frac{d}{dx} (\sin x) = \cos x \)
• \( \frac{d}{dx} (\cos x) = -\sin x \)
• \( \frac{d}{dx} (\sec x) = \sec x \tan x \)

Differentiation helps identify important features of functions, such as critical points, maxima, and minima by analyzing where the derivative is zero or undefined.
These features can be used when applying the Extreme Value Theorem.

Trigonometric Functions

Trigonometric functions are a set of functions that relate the angles of a triangle to the lengths of its sides. Common trigonometric functions include sine, cosine, and tangent, while secant, cosecant, and cotangent are their reciprocal functions. In this exercise, the focus is on the secant function, \( \sec x \), which is the reciprocal of cosine:

• \( \sec x = \frac{1}{\cos x} \)

The secant function is periodic and exhibits vertical asymptotes where the cosine function is zero (i.e., where \( \cos x = 0 \)). The behavior of trigonometric functions is complex because they oscillate between specific values, sometimes leading to undefined points.
Understanding the behavior of these functions is pivotal when analyzing critical points, particularly in periodic intervals, as you'll see in the exercise above.

Critical Points

Critical points are values of \( x \) where the derivative of a function is either zero or undefined. These points are significant because they may represent local maxima, minima, or points of inflection. However, not all critical points result in extremes. In the given exercise, the function

• \( g'(x) = \sec x \tan x \)

does not equal zero for any \( x \) in the interval \([-\frac{\pi}{6}, \frac{\pi}{3}]\). Because \( \sec x \) and \( \tan x \) cannot be zero for any real number \( x \) within this interval, there are no critical points.
This is important because, when seeking absolute extremes on a closed interval, you need to only consider the endpoints and any potential critical points.
Without critical points in this case, evaluating the function at the endpoints determines the absolute extrema on the interval.