Problem 40 - Thomas Calculus

Question

In Exercises \(37-40 :\) a. Find the local extrema of each function on the given interval, and say where they are assumed. 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}

Step-by-Step Solution

Answer

Derivatives provide critical points and indicate whether the function is increasing or decreasing for identifying local extrema. Use graphing for visual analysis.

Step 1: Find the Derivative of f(x)

Given the function \( f(x) = \sec^{2}x - 2\tan x \), we need to find its derivative \( f'(x) \).- Use the derivative of \( \sec x \) which is \( \sec x \tan x \) and the derivative of \( \tan x \) which is \( \sec^2 x \).- The derivative \( f'(x) \) is given by: \[\begin{align*}f'(x) &= \frac{d}{dx}(\sec^2 x) - 2\frac{d}{dx}(\tan x) \ &= 2\sec^2 x \cdot \sec x \tan x - 2\sec^2 x \ &= 2\sec^3 x \tan x - 2\sec^2 x.\end{align*}\]

Step 2: Determine Critical Points

Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined.- Set \( f'(x) = 0 \): \[2\sec^3 x \tan x - 2\sec^2 x = 0\]Factor out \( 2\sec^2 x \):- \[ 2\sec^2 x (\sec x \tan x - 1) = 0 \]Solutions are \( \sec x \tan x = 1 \). Find \( x \) values that satisfy these solutions in the interval \(-\pi/2 < x < \pi/2\).

Step 3: Solve for x within the Interval

From \( \sec x \tan x = 1 \), substitute \( \sec x = \frac{1}{\cos x} \) and \( \tan x = \frac{\sin x}{\cos x} \):- \( \frac{\sin x}{\cos^2 x} = 1 \) implies \( \sin x = \cos^2 x \).Solve this equation for \( x \) within the interval \(-\pi/2 < x < \pi/2\). This typically involves trigonometric identities or solving the resulting equations.

Step 4: Analyze the Sign of f'(x)

To analyze when \( f(x) \) is increasing or decreasing:- Identify intervals where \( f'(x) > 0 \) (function increasing) and \( f'(x) < 0 \) (function decreasing).Determine these intervals using test points and the sign of \( 2\sec^3 x \tan x - 2\sec^2 x \). This helps in identifying local maxima or minima.

Step 5: Evaluate f(x) at Critical Points

Calculate \( f(x) \) at any critical points found to determine the local extrema.Evaluate the function at these critical points to find the local maximum or minimum values.

Step 6: Sketch the Graph of f(x) and f'(x)

Using a graphing tool or software, plot both \( f(x) = \sec^2 x - 2\tan x \) and its derivative \( f'(x) = 2\sec^3 x \tan x - 2\sec^2 x \) over the interval \(-\pi/2 < x < \pi/2\).Comment on how the derivative \( f'(x) \) influences the shape and behavior of the graph of \( f(x) \). Notice how changes in sign of \( f'(x) \) correspond to local extrema in \( f(x) \).

Key Concepts

Local Extrema

Local extrema refer to the highest or lowest points in a specific region of a function’s graph. To find these points, we’re interested in when the slope of the function becomes zero or undefined, as these are the markers of potential turning points. For the given function, these occur at critical points where the derivative of the function, denoted as \( f'(x) \), equals zero or does not exist.

By solving \( f'(x) = 0 \), we find potential candidates for local minima or maxima. In this problem, once the derivative is found and simplified, setting it to zero allows us to find \( x \) values that mark these extrema, provided these values also lie within the interval \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \].

After determining these critical points, it is important to substitute them back into the original function, \( f(x) \), to calculate their corresponding \( y \)-values. These values help us confirm whether a point is indeed a local maximum or minimum by comparing the function's output in nearby areas.

Derivative Analysis

Analyzing the derivative of a function is essential for understanding its behavior, including identifying local extrema. Given the original function \( f(x) = \sec^2 x - 2\tan x \), the derivative \( f'(x) \), which is \( 2\sec^3 x \tan x - 2\sec^2 x \), tells us about the slope of \( f(x) \) at any given point.

Finding where \( f'(x) = 0 \) helps pinpoint critical points. These are the spots where changes in the function's direction might occur—shifting from increasing to decreasing or vice versa. To solve this, factor \( f'(x) \) and analyze its components within the given interval.

Analyzing where \( f'(x) \) is greater than zero indicates intervals where \( f(x) \) is increasing, while \( f'(x) \) being less than zero indicates decreasing intervals. Undetermined points in the derivative can also point to essential discontinuities or vertical tangents.

Trigonometric Functions

Trigonometric functions like \( \sec x \) and \( \tan x \) are crucial in calculus, mainly due to their repeating nature and distinctive characteristics. Understanding how they behave and their derivatives is vital. For \( f(x) = \sec^2 x - 2\tan x \), knowing that \( \sec x = \frac{1}{\cos x} \) and \( \tan x = \frac{\sin x}{\cos x} \) helps track how these change within an interval.

The derivatives of \( \sec x \) and \( \tan x \) as \( \sec x \tan x \) and \( \sec^2 x \) respectively illustrate how their rates of change are interconnected and lead to complex behaviors in combined functions.

In our problem, substituting known trigonometric identities into equations such as \( \sec x \tan x = 1 \) enables us to solve for \( x \), revealing meaningful insights about the function's behavior over different parts of the defined interval.

Graphing Techniques

Graphing functions requires understanding both the algebraic expressions and the behavior indicated by their derivatives. By plotting \( f(x) = \sec^2 x - 2\tan x \) and its derivative \( f'(x) = 2\sec^3 x \tan x - 2\sec^2 x \), we can visually identify where the function increases or decreases and where it has local extrema.

Graphing these not only provides a pictorial depiction of the function but also supports the algebraic analysis. For instance, spot checks against known critical points or where the derivative changes sign are reflected graphically.

This allows for confirming intuitive insights: local maximums and minimums are where the graph changes direction, corresponding to the derivative's sign shift. Moreover, clear graphs help to correlate and comprehend the fundamental relationships and influences between \( f(x) \) and \( f'(x) \), sharpening our intuition about the nature of trigonometric functions in calculus.