Problem 65 - University Calculus: Early Transcendentals

Question

a. The function \(y=\cot x-\sqrt{2} \csc x\) has an absolute maximum value on the interval \(0

Step-by-Step Solution

Answer

The absolute maximum value is -1, at \(x=\frac{\pi}{4}\).

Step 1: Understand the Function and Interval

We need to find the absolute maximum value of the function \(y=\cot x-\sqrt{2} \csc x\) on the interval \((0, \pi)\). The function involves trigonometric functions cotangent and cosecant, and we must consider their behavior within the given interval.

Step 2: Find the Derivative

To find critical points, we first need the derivative of \(y\). The derivative of \(\cot x\) is \(-\csc^2 x\), and the derivative of \(\sqrt{2} \csc x\) is \(-\sqrt{2} \csc x \cot x\). Thus, the derivative is:\[\frac{dy}{dx} = -\csc^2 x + \sqrt{2} \csc x \cot x.\]

Step 3: Set Derivative to Zero

Set the derivative equal to zero to find critical points:\[-\csc^2 x + \sqrt{2} \csc x \cot x = 0.\]Factor out \(\csc x\):\[\csc x(-\csc x + \sqrt{2} \cot x) = 0.\]Since \(\csc x eq 0\) (it would imply \(x = n\pi\), which is not in the open interval \((0, \pi)\)), solve:\[-\csc x + \sqrt{2} \cot x = 0.\]

Step 4: Solve for Critical Points

Simplify the equation for critical points:\[\sqrt{2} \cot x = \csc x \Rightarrow \sqrt{2} \frac{\cos x}{\sin x} = \frac{1}{\sin x}.\]This simplifies to:\[\sqrt{2} \cos x = 1,\]\[\cos x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.\]The solutions for \(x\) in \((0, \pi)\) are \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).

Step 5: Evaluate Function at Critical Points

Calculate \(y\) at \(x = \frac{\pi}{4}\):\[y = \cot \frac{\pi}{4} - \sqrt{2} \csc \frac{\pi}{4} = 1 - \sqrt{2} \cdot \sqrt{2} = 1 - 2 = -1.\]Calculate \(y\) at \(x = \frac{3\pi}{4}\):\[y = \cot \frac{3\pi}{4} - \sqrt{2} \csc \frac{3\pi}{4} = -1 - \sqrt{2} \cdot \sqrt{2} = -1 - 2 = -3.\]

Step 6: Determine the Maximum Value

Compare the computed values: \(-1\) and \(-3\). The function's value at \(x=\frac{\pi}{4}\) is greater than at \(x=\frac{3\pi}{4}\). Therefore, the absolute maximum value of the function \(y\) on \((0, \pi)\) is \(-1\).

Step 7: Graph and Verify

Graph the function \(y=\cot x-\sqrt{2} \csc x\) on a calculator or software to visually identify its behavior over the interval \((0, \pi)\). The graph will confirm the function has an absolute maximum at \(x=\frac{\pi}{4}\) with a value of \(-1\), as shown by a peak in the graph.

Key Concepts

Absolute Maximum

In mathematics, finding the absolute maximum of a function involves identifying the highest point on an interval. It shows the greatest value the function attains during this interval. When working on trigonometric functions like the one in this exercise, the task involves considering all critical points and endpoints within the interval. Since this interval is \(0, \pi\), the focus is purely on critical points within this range.

**Why is it important?** Understanding the absolute maximum is essential for various applications:

• It helps optimize problems, especially in physics and engineering.
• In economics, it identifies the highest potential profit on a curve.
• It provides a clear view of the peak performance of systems.

This concept arranges and prioritizes critical and boundary points to predict the maximum value reliably. Here, we found a maximum at \(x = \frac{\pi}{4}\) with a function value of \(-1\). This point isn’t just any peak, but the highest peak achievable within the given range.

Derivative

The derivative of a function plays a crucial role in calculus. It provides the rate at which the function value changes as the input varies. Specifically, when we deal with functions like \(y=\cot x-\sqrt{2} \csc x\), the derivative helps us avoid complex manual calculations.

The derivative, noted as \(\frac{dy}{dx}\), evaluates to:

• \(-\csc^2 x + \sqrt{2} \csc x \cot x\)

This expression is derived by differentiating each term separately, using known derivatives of trigonometric functions:

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

The role of the derivative in this context includes:

• Finding critical points to determine maximas and minimas.
• Assessing where the function increases or decreases.

By setting \(\frac{dy}{dx} = 0\), we identify where changes cease, leading us to these critical points.

Critical Points

Critical points occur where the derivative of a function equals zero or is undefined. They are integral in finding where maxima, minima, or inflection points might occur. In our task for the function \(y=\cot x-\sqrt{2}\csc x\), the critical points define potential peaks and troughs between \(0\) and \(\pi\).

**Solving for Critical Points**
To find them, we set the derivative \(\frac{dy}{dx} \,=\, -\csc^2 x + \sqrt{2} \csc x \cot x \,=\, 0\) and solve for \(x\):

• Factor and simplify the equation.
• Lead to \(\sqrt{2} \cos x = 1\) which simplifies to \(\cos x = \frac{1}{\sqrt{2}}\).
• This results in critical points at \(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\).

Evaluating these points allows us to determine which achieves the desired extremum (maximum or minimum). In this exercise, comparison of function values at these points established that the absolute maximum occurs at \(x=\frac{\pi}{4}\) with \(y=-1\). Recognizing these critical places aids efficiency, letting us focus our final assessments on fewer, more likely locations.