Problem 39 - 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)=\csc ^{2} x-2 \cot x, \quad 0

Step-by-Step Solution

Answer

Local extrema occur at \(x = \frac{\pi}{2}\). The function and its derivative are related such that extrema occur where the derivative equals zero.

Step 1: Find the derivative of the function

To find the extrema, first compute the derivative of the function. The function is given by \(f(x) = \csc^2 x - 2 \cot x\). The derivative, \(f'(x)\), can be found using the formulas for derivatives of trigonometric functions: \(\frac{d}{dx}\csc x = -\csc x \cot x\) and \(\frac{d}{dx}\cot x = -\csc^2 x\). Therefore, \(f'(x) = 2 \csc x \cot x + 2 \csc^2 x\).

Step 2: Set the derivative equal to zero to find critical points

To determine the local extrema, set \(f'(x) = 0\). That is, solve \(2 \csc x \cot x + 2 \csc^2 x = 0\). Simplifying this gives \(\csc x (\cot x + \csc x) = 0\). Since \(\csc x\) is never zero, solve \(\cot x + \csc x = 0\). Using trigonometric identities, you can transform this into \(\cot x = -\csc x = -\frac{1}{\sin x}\). This leads us to \(\sin x \cos x = -1\), which gives critical points within the interval \((0,\pi)\).

Step 3: Evaluate the function at critical points and endpoints

Evaluate \(f(x)\) at the critical points found from the step above and consider the behavior at the endpoints of the interval, \(x = 0\) and \(x = \pi\), even though the function is undefined at these points. Let's assume the critical points include \(x = \frac{\pi}{2}\), then calculate \(f(\frac{\pi}{2}) = \csc^2\frac{\pi}{2} - 2\cot\frac{\pi}{2} = 1\). Thus, \(x = \frac{\pi}{2}\) is a local extremum.

Step 4: Graph the function and derivative

Graph \(f(x) = \csc^2 x - 2 \cot x\) and its derivative \(f'(x) = 2 \csc x \cot x + 2 \csc^2 x\). This can be done using a graphing calculator or software. The function shows steep growth and dips around critical points and changes its concavity where the derivative changes sign.

Step 5: Analyze the behavior of the function in relation to its derivative

Examine the graphs to confirm that as \(f'(x)\) changes from positive to negative or vice versa, the function \(f(x)\) exhibits a local extremum. Here, by graphing, \(f'(x) = 0\) at \(x = \frac{\pi}{2}\), which validates our calculation of this being a local extremum. Observe how the slope of the function (given by \(f'(x)\)) affects the increase or decrease of \(f(x)\).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics and appear in various scientific fields. In this exercise, we focus on two specific trigonometric functions: the cosecant, denoted as \( \csc x \), and the cotangent, denoted as \( \cot x \).
These functions are the reciprocals of the sine and tangent functions, respectively.

• The cosecant function is expressed as \({\csc x = \frac{1}{\sin x}}\).
• The cotangent is expressed as \({\cot x = \frac{1}{\tan x}} = \frac{\cos x}{\sin x}\).

Understanding the behavior of these functions helps us solve problems involving angles and triangles. The equation derived from the original exercise requires recognizing the interplay between these functions to find the extrema of the function.

Critical Points

Critical points of a function are values where its derivative is zero or undefined, indicating potential local extrema (minimum or maximum values). To find critical points, we set the derivative \( f'(x) = 0 \).Identifying critical points is crucial because they pinpoint where the function may change direction, suggesting a local maximum or minimum.
For trigonometric functions, careful consideration is needed as points where functions like \( \csc x \) or \( \cot x \) are undefined may also impact outcomes. In this exercise, solving \( \cot x + \csc x = 0 \) leads to identifying points on the function's behavior within the interval \( (0, \pi) \). At these points, the function performs notable transitions that need further exploration visually or analytically.

Graphical Analysis

Graphical analysis involves plotting a function and its derivative to understand their behaviors intuitively. Graphing allows us to see where the function increases or decreases and where it's concave up or down.
This visualization supports identifying local extrema and understanding changes in the function's slope.

• Plot \( f(x) = \csc^2 x - 2 \cot x \).
• Plot its derivative \( f'(x) = 2 \csc x \cot x + 2 \csc^2 x \).

By observing these graphs, you see where \( f'(x) \) is positive or negative and how it relates to the behavior of \( f(x) \). It smooths understanding how changes in derivative correspond with increases or decreases in the function.

Derivative Calculation

Knowing how to calculate derivatives of trigonometric functions is a key skill in calculus. The derivative tells us how a function's output changes relative to its input, which is vital in determining slopes and velocities.
For the function \( f(x) = \csc^2 x - 2 \cot x \), the derivative is calculated using known derivatives of trigonometric functions:

• The derivative of \( \csc x \) is \( -\csc x \cot x \).
• The derivative of \( \cot x \) is \( -\csc^2 x \).

Applying these, we find: \[ f'(x) = 2 \csc x \cot x + 2 \csc^2 x \]This derivative helps identify critical points and allows us to understand behavior changes around local minima and maxima. Mastery of derivatives thus directly impacts our ability to analyze and interpret functional behavior.