Problem 32 - Thomas Calculus

Question

In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ f(\theta)=\tan \theta, \quad-\frac{\pi}{3} \leq \theta \leq \frac{\pi}{4} $$

Step-by-Step Solution

Answer

Absolute minimum is at \((-\frac{\pi}{3}, -\sqrt{3})\); absolute maximum is at \((\frac{\pi}{4}, 1)\).

Step 1: Determine Critical Points

To find the critical points of the function on the given interval, calculate the derivative of the function. The derivative of \( f(\theta) = \tan \theta \) is \( f'(\theta) = \sec^2 \theta \). Critical points occur where \( f'(\theta) = 0 \) or \( f'(\theta) \) does not exist. Here, \( \sec^2 \theta \) is never zero, but it is undefined where \( \theta = \frac{\pi}{2} + k\pi \), which is outside the interval \( -\frac{\pi}{3} \leq \theta \leq \frac{\pi}{4} \). Thus, there are no critical points within the interval.

Step 2: Evaluate the Function at Endpoints

The absolute extrema of a continuous function on a closed interval can occur at endpoints or critical points. Since there are no critical points in the interval, evaluate \( f(\theta) \) at the endpoints. Compute \( f\left(-\frac{\pi}{3}\right) = \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \) and \( f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \).

Step 3: Compare Values to Find Absolute Extrema

Compare the computed values to determine the absolute extrema. The value \( -\sqrt{3} \) at \( \theta = -\frac{\pi}{3} \) is the absolute minimum, and the value \( 1 \) at \( \theta = \frac{\pi}{4} \) is the absolute maximum on the interval.

Step 4: Graph the Function and Identify Extrema Points

Graph \( f(\theta) = \tan \theta \) over the interval \(-\frac{\pi}{3} \leq \theta \leq \frac{\pi}{4}\). The absolute minimum occurs at \( (-\frac{\pi}{3}, -\sqrt{3}) \) and the absolute maximum occurs at \( (\frac{\pi}{4}, 1) \). Mark these points on the graph to clearly show where the extrema occur.

Key Concepts

Critical Points Analysis

When analyzing functions for extremum (maximum or minimum values), a crucial step is to determine the critical points. These are the points where the function's derivative is either zero or does not exist. In the context of the original exercise, we focus on the function \( f(\theta) = \tan \theta \). The derivative, \( f'(\theta) = \sec^2 \theta \), highlights potential critical points.
Critical points help in identifying where the maximum or minimum values might occur within a given interval. In our scenario, \( \sec^2 \theta \) is never zero over the interval \( -\frac{\pi}{3} \leq \theta \leq \frac{\pi}{4} \).

• Critical points occur where the derivative is zero or undefined.
• Here the derivative is undefined at points like \( \theta = \frac{\pi}{2} + k\pi \), but these lie outside the interval of consideration.

This means we do not find any critical points within the given boundaries. Knowing this helps us redirect our focus towards the endpoints for analysis of extrema.

Derivative Calculation

Calculating the derivative is a foundational step in exploring extrema for calculus problems. Given the function \( f(\theta) = \tan \theta \), we differentiate to obtain \( f'(\theta) = \sec^2 \theta \). This derivative provides insight into the slope of the tangent line to the function at various points.
Derivative calculation lets us observe rate changes and potential regions where extrema could occur. Since \( \sec^2 \theta \) does not simplify to zero, this absence of zero-derivative points forces us to look elsewhere. Here's a quick look at the procedures:

• Differentiating \( \tan \theta \) leads to \( \sec^2 \theta \).
• The derivative is crucial for analyzing the function's behavior.

In this case, because the derivative is not zero anywhere in the interval, identifying extrema involves evaluating the function at the interval's edges.

Function Interval Evaluation

Once critical points and derivatives point out that they offer no extrema guidance, we turn our attention to evaluating the function at the interval’s boundaries. In this case, these boundaries are \( -\frac{\pi}{3} \) and \( \frac{\pi}{4} \). Evaluating \( f(\theta) = \tan \theta \) at these specific values involves straightforward trigonometric computations.
Let’s break down the evaluation process:

• Calculate \( f\left(-\frac{\pi}{3}\right) = \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \).
• Compute \( f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \).

The results show that the absolute minimum value is \( -\sqrt{3} \) at \( \theta = -\frac{\pi}{3} \) and the absolute maximum value is \( 1 \) at \( \theta = \frac{\pi}{4} \). By evaluating only at the boundaries, we successfully identify where the extrema occur on our given interval.