Problem 60 - Thomas Calculus

Question

a. Find the local extrema of each function on the given interval, and say where they occur. b. Graph the function and its derivative together. Comment on the behavior of \(f\) in relation to the signs and values of \(f^{\prime}. \) $$ f(x)=\sec ^{2} x-2 \tan x, \quad \frac{-\pi}{2}< x <\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The local minimum of \( f(x) \) occurs at \( x = \frac{\pi}{4} \).

Step 1: Identify the Function and Interval

We are given the function \( f(x) = \sec^2 x - 2 \tan x \) and need to find its local extrema in the interval \( -\frac{\pi}{2} < x < \frac{\pi}{2} \). The function involves trigonometric identities, which are crucial to understand its derivative.

Step 2: Find the Derivative of the Function

To find the local extrema, we first need to determine the derivative \( f'(x) \). We start with the derivatives of \( \sec^2 x \) and \( \tan x \). The derivative of \( \sec^2 x \) is \( 2 \sec^2 x \tan x \) and of \( \tan x \) is \( \sec^2 x \). Combining these, we have: \[ f'(x) = \frac{d}{dx}(\sec^2 x - 2 \tan x) = 2 \sec^2 x \tan x - 2 \sec^2 x \] Simplifying, \( f'(x) = 2 \sec^2 x (\tan x - 1) \).

Step 3: Set the Derivative to Zero to Find Critical Points

To locate critical points, set \( f'(x) = 0 \): \[ 2 \sec^2 x (\tan x - 1) = 0 \] For this to be true, either \( \sec^2 x eq 0 \) or \( \tan x - 1 = 0 \), so: \( \tan x = 1 \). Solving this gives \( x = \frac{\pi}{4} \), within the interval.

Step 4: Determine Local Extrema Using Second Derivative Test

For confirmation, use the second derivative test. The second derivative is:\[ f''(x) = 2(\sec^2 x (\sec^2 x - 2)) \sec^2 x - 2 \sec^2 x \tan x \] Evaluate \( f''(x) \) at \( x = \frac{\pi}{4} \). If \( f''(x) > 0 \), the point is a local minimum; if \( f''(x) < 0 \), it's a local maximum. After calculations, \( f''(x) > 0 \) at \( x = \frac{\pi}{4} \), indicating a local minimum.

Step 5: Graph the Function and Its Derivative

To understand behavior, graph both \( f(x) = \sec^2 x - 2 \tan x \) and \( f'(x) \). Observe that where \( f'(x) \) changes sign, \( f(x) \) transitions from increasing to decreasing or vice versa. At \( x = \frac{\pi}{4} \), \( f'(x) \) changes from positive to negative, corroborating a local minimum position.

Step 6: Comment on the Behavior of f in Relation to f'

The graph of \( f \) and its derivative \( f' \) reveals that when \( f'(x) > 0 \), \( f(x) \) is increasing, and when \( f'(x) < 0 \), \( f(x) \) is decreasing. The critical point at \( x = \frac{\pi}{4} \) corresponds to a minimum where \( f'(x) = 0 \). The signs of \( f'(x) \) determine the increasing or decreasing behavior of \( f(x) \).

Key Concepts

Critical Points

Critical points are vital in determining where a function's slope changes, which helps locate local maxima or minima. For a function like the one given by \[f(x) = \sec^2 x - 2 \tan x\]these are points where the derivative \[f'(x)\]equals zero or is undefined. Identifying critical points requires that we find when the derivative equals zero or does not exist. In our example, the derivative simplifies to:\[f'(x) = 2 \sec^2 x (\tan x - 1)\]The critical points occur when \[\tan x = 1\]because \[\sec^2 x\]never equals zero. Solving gives \[x = \frac{\pi}{4}\] within the specified interval, which is the location for potential extrema. Understanding critical points provides the foundation for locating and classifying these important features in the graph of a function.

Derivative Test

The derivative test is a primary tool to identify the nature of critical points, be it a local maximum, minimum, or neither. After finding where the derivative \(f'(x)\)equals zero, the next step is to determine whether these are indeed points of local extremum. The test typically involves evaluating the second derivative:- **Second Derivative Test**: If \(f''(x) > 0\)at a critical point, the function has a local minimum there. Conversely, if \(f''(x) < 0\),it’s a local maximum.In our exercise, we have determined \(x = \frac{\pi}{4}\) as a critical point. Analyzing the second derivative confirms it is:\[f''(x) = 2(\sec^2 x (\sec^2 x - 2)) \sec^2 x - 2 \sec^2 x \tan x\]Evaluating at \(x = \frac{\pi}{4}\), if \(f''(x) > 0\), it indicates a local minimum. This process helps confirm whether these critical points found are indeed extrema and what kind they are.

Trigonometric Functions

Trigonometric functions are integral in analyzing the given function \( f(x) = \sec^2 x - 2 \tan x \). Their properties affect the function’s shape and behavior due to the periodicity, asymptotes, and range. In this problem:

• The secant function, \(\sec x\), is the reciprocal of the cosine function and is undefined where \(\cos x = 0\).
• The tangent function, \(\tan x\), relates to the sine and cosine functions \((\tan x = \frac{\sin x}{\cos x})\) and is also undefined when \(\cos x = 0\).

When evaluating derivatives, recognizing these properties helps simplify expressions and concentrate on critical behaviors at specific points like \(x = \frac{\pi}{4}\).This understanding helps in predicting changes and identifying critical points as the function trends.

Interval Analysis

Interval analysis assesses how the function behaves over a specific range. In our exercise, we are looking at the interval \(-\frac{\pi}{2} < x < \frac{\pi}{2}\). This range governs the behavior and limits of both first and second derivatives:

• Analyze when \(f'(x) > 0\): Function is increasing.
• Analyze when \(f'(x) < 0\): Function is decreasing.

Points at which \(f'(x) = 0\) or is undefined within this interval signify possible local maxima or minima.Performing an interval analysis highlights sections where potential changes occur in the slope, aiding in graphing the function accurately and stating where these local extremum may exist. Understanding the onset of increases, decreases, or points of concavity helps sketch both the function and its derivative effectively, revealing these characteristics visually.