Problem 34 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find 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 function \(g(x) = \sec x\) on the interval \([- \frac{\pi}{6}, \frac{\pi}{3}]\) has an absolute maximum of \(2 \sqrt{3}\) at \(x = -\frac{\pi}{6}\) and an absolute minimum of \(2\) at \(x = \frac{\pi}{3}\).

Step 1: Find the derivative

Take the derivative of \(g(x) = \sec x\) to find the function's critical points. The derivative of secant function \(g'(x) = \sec x \cdot \tan x\).

Step 2: Find the critical points

Set the derivative \(g'(x)\) equivalent to zero and solve for \(x\). However, for the secant function, there is no real number which makes \(\sec x \cdot \tan x\) equal to zero, so there are no critical points in this interval.

Step 3: Evaluate the function at the endpoints

Evaluate the given function at the endpoints of the interval provided, i.e., at \(x = - \frac{\pi}{6}\) and \(x = \frac{\pi}{3}\). After evaluating, find that \(g(- \frac{\pi}{6}) = 2 \sqrt{3}\) and \(g(\frac{\pi}{3}) = 2\).

Step 4: Identify the absolute extrema

There are no critical points, so the maximum and minimum must occur at the endpoints. Comparing the values, \(2 \sqrt{3}\) is greater than \(2\), hence \(g(x)\) has an absolute maximum of \(2 \sqrt{3}\) at \(x = -\frac{\pi}{6}\) and an absolute minimum of \(2\) at \(x = \frac{\pi}{3}\).

Key Concepts

Derivative

In calculus, the derivative is a fundamental concept used to determine the rate at which a function is changing at any given point. It provides us with invaluable insights into the behavior of functions.
For the function given, which is the secant function, the first step in solving for extrema involves finding the derivative. The derivative of the secant function, noted as \(g'(x)\), is \(\sec x \cdot \tan x\).
This derivative tells us how the secant function behaves as \(x\) changes. Derivatives can help identify points where the slope of the tangent is zero, which are the critical points.

• For a function \(f(x)\), the derivative is often noted as \(f'(x)\).
• Derivatives are used to find increasing or decreasing behavior of functions.
• They are crucial in finding critical points, which might be candidates for extrema.

Secant Function

The secant function, written as \(\sec x\), is one of the six fundamental trigonometric functions. It is the reciprocal of the cosine function, meaning \(\sec x = \frac{1}{\cos x}\). Due to its nature, the secant function has certain characteristics and restrictions.
Unlike the cosine function, the secant function has undefined points where the cosine of \(x\) equals zero. These occur at odd multiples of \(\frac{\pi}{2}\).

• The secant function is periodic and repeats every \(2\pi\).
• It has vertical asymptotes where the cosine function equals zero.
• Understanding secant is important for analyzing and graphing trigonometric functions.

It is essential when evaluating \(\sec x\) to ensure you're in intervals where the cosine is not zero to avoid undefined values.

Critical Points

Critical points are points in the function's domain where the derivative is zero or undefined. They are essential when exploring the behavior of the function and determining potential local maxima or minima.
In our case, the function \(g(x) = \sec x\) has the derivative \( g'(x) = \sec x \cdot \tan x \). To find critical points, we would set this derivative to zero, resolving \(\tan x = 0\) because \(\sec x\) is never zero.

• For \(\sec x \cdot \tan x = 0\), it does not happen for our range since \(\sec x\) is not zero and \(\tan x = 0\) corresponds to integral multiples of \(\pi\), which are not within the given interval.
• Without critical points inside the interval, extrema must happen at the interval’s endpoints.
• This step is crucial to ascertain if there are points in the middle of the interval to consider.

Absolute Extrema

Absolute extrema refer to the highest or lowest values that a function can attain over a closed interval. For the function \(g(x) = \sec x\) within the interval \([-\frac{\pi}{6}, \frac{\pi}{3}]\), we assess its values at the boundaries of this interval since there are no critical points.
To find the absolute extrema, evaluate the secant function at the endpoint values:

• At \(x = -\frac{\pi}{6}\), \(g(-\frac{\pi}{6}) = 2\sqrt{3}\). Here, the secant function reaches its maximum within the specified range.
• At \(x = \frac{\pi}{3}\), \(g(\frac{\pi}{3}) = 2\), which presents the minimum value in the range, given the absence of values exceeding or below it.

Without critical points in this closed interval, the maxima and minima naturally reside at the start or end points of the interval, illustrating how absolute extrema are determined in such cases.