Problem 59 - Thomas Calculus in SI Units

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

Step-by-Step Solution

Answer

Local minimum at \( x = \pi \) with a function value of 1.

Step 1: Find the Derivative

First, find the derivative of the given function. For the function \( f(x) = \csc^2(x) - 2\cot(x) \), using the derivatives \( \frac{d}{dx}\csc(x) = -\csc(x)\cot(x) \) and \( \frac{d}{dx}\cot(x) = -\csc^2(x) \), we find:\[f'(x) = -2\csc(x)\cot(x) + 2\csc^2(x)\]

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero and solve for \( x \):\[-2\csc(x)\cot(x) + 2\csc^2(x) = 0\]Simplifying gives:\[2\csc(x)(\csc(x) - \cot(x)) = 0\]Since \( \csc(x) eq 0 \), solve:\[\csc(x) = \cot(x) \]

Step 3: Solve \( \csc(x) = \cot(x) \)

Recall that \( \csc(x) = \frac{1}{\sin(x)} \) and \( \cot(x) = \frac{\cos(x)}{\sin(x)} \), so:\[\frac{1}{\sin(x)} = \frac{\cos(x)}{\sin(x)} \]Simplify to find:\[1 = \cos(x) \]Thus, \( x = \frac{\pi}{2} \). Since \( \cos(x) = 1 \) never happens in \( 0 < x < \pi \), critical point is only at \( x = 0 \) or \( x = \pi \), not within the interval.

Step 4: Verify and Find Extrema

Evaluate the endpoints as the interval suggests no internal critical points:- At \( x \to 0^+ \), \( f(x) \to \infty \).- At \( x = \pi \), \( f(\pi) = 1 \).Therefore, there is no local maximum, but a local minimum of 1 occurs at \( x = \pi \).

Step 5: Graphical Interpretation

Graph \( f(x) = \csc^2(x) - 2\cot(x) \) and its derivative together. Notice that \( f'(x) > 0 \) in regions implying \( f(x) \) is increasing, and \( f'(x) < 0 \) where \( f(x) \) is decreasing. The behavior of \( f \) correlates with the sign of \( f'(x) \).

Key Concepts

Derivative Calculation

To solve problems involving local extrema, we often begin by calculating the derivative of the function in question. The derivative tells us how the function is changing at a specific point.
In simple terms, if we express a function as \( f(x) \), its derivative, denoted as \( f'(x) \), represents the rate of change of \( f \) with respect to \( x \).

Consider the function given in the exercise:

• \( f(x) = \csc^2(x) - 2\cot(x) \)

Here, we are dealing with the derivatives of trigonometric functions.

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

Using these, we find:

• \( f'(x) = -2\csc(x)\cot(x) + 2\csc^2(x) \)

This derivative helps us find critical points through further algebraic manipulation.

Trigonometric Functions

Understanding trigonometric functions is crucial when analyzing functions like \( f(x) = \csc^2(x) - 2\cot(x) \). These functions, such as cosecant (\( \csc \)) and cotangent (\( \cot \)), are derived from the basic sine and cosine functions.

• The cosecant is the reciprocal of sine: \( \csc(x) = \frac{1}{\sin(x)} \).
• Cotangent is the ratio of cosine to sine: \( \cot(x) = \frac{\cos(x)}{\sin(x)} \).

Recognizing these relationships is vital when solving trigonometric equations. In our case, to solve \( \csc(x) = \cot(x) \):

• Rewrite as \( \frac{1}{\sin(x)} = \frac{\cos(x)}{\sin(x)} \).
• Simplify to \( 1 = \cos(x) \).

However, within the interval \( 0 < x < \pi \), this condition is unmet, guiding us to investigate only the boundaries of the interval.

Graphical Analysis

Graphical analysis provides valuable insight into the behavior of functions and their derivatives. By examining the graph of \( f(x) = \csc^2(x) - 2\cot(x) \) alongside its derivative \( f'(x) \), we can visually interpret the function's behavior.
When graphing, notice:

• Where \( f'(x) > 0 \), the original function \( f(x) \) is increasing.
• Where \( f'(x) < 0 \), \( f(x) \) is decreasing.

This information indicates possible local maxima or minima. In the step-wise solution, \( f(x) \) has a local minimum at the endpoint \( x = \pi \).
The endpoint assessment and graphs reveal that \( f(x) \) trends towards infinity as \( x \) approaches \( 0^+ \), pinpointing where the function does not yield a local maximum inside the set interval. Graphs can thus clarify the conclusions drawn from purely algebraic investigations.