Problem 60 - University Calculus: Early Transcendentals

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)=-2 x+\tan x, \quad \frac{-\pi}{2}

Step-by-Step Solution

Answer

Local maximum at \(x = -\frac{\pi}{4}\) and local minimum at \(x = \frac{\pi}{4}\) within the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\) for \(f(x) = -2x + \tan x\).

Step 1: Find the Derivative

To find the derivative of the function, we'll use the derivative rules. The derivative of \(-2x\) is \(-2\), and the derivative of \(\tan x\) is \(\sec^2 x\). Therefore, the derivative of \(f(x) = -2x + \tan x\) is given by: \[ f'(x) = -2 + \sec^2 x. \]

Step 2: Find Critical Points

Critical points occur where the derivative equals zero or is undefined. Therefore, we solve the equation \(-2 + \sec^2 x = 0\). Rearrange this to \(\sec^2 x = 2\), which implies \(\cos^2 x = \frac{1}{2}\). Solving \(\cos x = \pm \frac{1}{\sqrt{2}}\) gives potential critical points at \(x = \pm \frac{\pi}{4}\). Since we are considering the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), both points are valid.

Step 3: Determine Local Extrema

To determine if the critical points are minima or maxima, evaluate the second derivative, or use the first derivative test. Here, we use the first derivative test:- For \(x < -\frac{\pi}{4}\), choose \(x = -\frac{\pi}{3}\), and find \(f'\) is positive.- For \(-\frac{\pi}{4} < x < \frac{\pi}{4}\), choose \(x = 0\), and find \(f'\) is negative.- For \(x > \frac{\pi}{4}\), choose \(x = \frac{\pi}{3}\), and again \(f'\) is positive. Thus, \(x = -\frac{\pi}{4}\) is a local maximum, and \(x = \frac{\pi}{4}\) is a local minimum.

Step 4: Graph the Function and its Derivative

Graph both \(f(x) = -2x + \tan x\) and its derivative \(f'(x) = -2 + \sec^2 x\). Notice the following:- \(f'(x) > 0\) in intervals \((-\frac{\pi}{2}, -\frac{\pi}{4})\) and \((\frac{\pi}{4}, \frac{\pi}{2})\), indicating \(f(x)\) is increasing in these intervals.- \(f'(x) < 0\) in \((-\frac{\pi}{4}, \frac{\pi}{4})\), indicating \(f(x)\) is decreasing there. The critical points are where \(f'(x) = 0\), signaling potential changes in direction.

Step 5: Analyze the Behavior of f with f'

From the graph and calculations, observe these behaviors:- At \(x = -\frac{\pi}{4}\), \(f'(x)\) changes from positive to negative, confirming a local maximum.- At \(x = \frac{\pi}{4}\), \(f'(x)\) changes from negative to positive, confirming a local minimum.Thus, the signs and transitions of \(f'\) align with standard behavior for identifying extrema.

Key Concepts

Derivative

The concept of a derivative is pivotal in calculus and helps us understand the rate at which a function is changing at any given point. When you look at a function like \( f(x) = -2x + \tan x \), you can determine its instantaneous rate of change by finding its derivative.

To do this, we apply derivative rules: the derivative of \(-2x\) is \(-2\), because it's a linear function with a constant slope, and the derivative of \( \tan x \) is \( \sec^2 x \), as derived from trigonometric identities. Therefore, the derivative of the whole function \( f(x) = -2x + \tan x \) is represented as \( f'(x) = -2 + \sec^2 x \).

Calculating the derivative helps identify how the function is behaving—whether it's increasing or decreasing at various points—which is critical for finding local extrema.

Critical Points

Critical points are where the derivative of a function is either zero or does not exist. These points are important because they help identify where the function might have local maxima or minima.

For the function \( f(x) = -2x + \tan x \), we set its derivative \( f'(x) = -2 + \sec^2 x \) to zero to find these critical points. Solving \( \sec^2 x = 2 \) leads to \( \cos^2 x = \frac{1}{2} \), yielding potential critical points at \( x = \pm \frac{\pi}{4} \), both of which lie within the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

These critical points are where the derivative changes signs, indicating a possible switch from increasing to decreasing (or vice versa) in the function's behavior.

First Derivative Test

The first derivative test is a method used to classify critical points as local maxima or minima by observing the sign changes of the derivative \( f'(x) \).

For instance, after identifying critical points at \( x = \pm \frac{\pi}{4} \) for the function \( f(x) = -2x + \tan x \), we evaluate the derivative in the intervals around these points:

• For \( x < -\frac{\pi}{4} \), \( f'(x) \) is positive, indicating that the function is increasing.
• Between \( -\frac{\pi}{4} < x < \frac{\pi}{4} \), \( f'(x) \) is negative, so the function is decreasing.
• For \( x > \frac{\pi}{4} \), \( f'(x) \) reverts to positive, showing an increasing function.

This pattern shows that at \( x = -\frac{\pi}{4} \), the derivative changes from positive to negative, signifying a local maximum. Conversely, at \( x = \frac{\pi}{4} \), it shifts from negative to positive, which points to a local minimum.

Using the first derivative test simplifies the process of determining the nature of these critical points.