Problem 8 - Calculus: Early Transcendentals

Question

Find all critical points and identify them as local maximum points, local minimum points, or neither. $$ y=\cos (2 x)-x $$

Step-by-Step Solution

Answer

Critical points are at \( x = \frac{7\pi}{12} + k\pi \) (local max) and \( x = \frac{11\pi}{12} + k\pi \) (local min), where \( k \) is an integer.

Step 1: Find the First Derivative

To find the critical points, we need the derivative of the function \( y = \cos(2x) - x \). Use the derivative rules for cosine and linear functions. The derivative of \( \cos(2x) \) is \(-2\sin(2x)\) (using the chain rule), and the derivative of \(-x\) is \(-1\). Therefore, the first derivative is: \[ y' = -2\sin(2x) - 1 \]

Step 2: Set the First Derivative to Zero

To find the critical points, set the first derivative equal to zero: \[ -2\sin(2x) - 1 = 0 \]Solve for \( \sin(2x) \) to find the critical points: \[ \sin(2x) = -\frac{1}{2} \]

Step 3: Solve for x

The equation \( \sin(2x) = -\frac{1}{2} \) implies that \( 2x \) must be at angles where sine is \(-\frac{1}{2}\). These angles are at \( 2x = \frac{7\pi}{6} + 2k\pi \) and \( 2x = \frac{11\pi}{6} + 2k\pi \), where \( k \) is an integer. Therefore, \( x \) can be solved as:\[ x = \frac{7\pi}{12} + k\pi \] \[ x = \frac{11\pi}{12} + k\pi \]

Step 4: Find the Second Derivative

Compute the second derivative to identify the nature of each critical point. The derivative of \(-2\sin(2x)\) is \(-4\cos(2x)\), so the second derivative is: \[ y'' = -4\cos(2x) \]

Step 5: Determine the Nature of Critical Points

For critical points \( x = \frac{7\pi}{12} + k\pi \) and \( x = \frac{11\pi}{12} + k\pi \), plug into the second derivative:- If \( y'' < 0 \), the point is a local maximum.- If \( y'' > 0 \), the point is a local minimum.For \( x = \frac{7\pi}{12} \), \( y''(\frac{7\pi}{12}) = -4\cos(\frac{7\pi}{6}) = -4(\frac{1}{2}) = -2 \), indicating a local maximum.For \( x = \frac{11\pi}{12} \), \( y''(\frac{11\pi}{12}) = -4\cos(\frac{11\pi}{6}) = -4(-\frac{1}{2}) = 2 \), indicating a local minimum.

Key Concepts

First Derivative Test

The First Derivative Test is a useful tool in calculus to determine where a function increases or decreases, and helps identify if a critical point is a local extremum. When analyzing a function like \( y = \cos(2x) - x \), the first step involves finding the derivative of the function. For example, the derivative of \( \cos(2x) \) is \(-2\sin(2x)\), and for \(-x\), it is \(-1\). Thus, the derivative \( y' = -2\sin(2x) - 1 \) helps us find where the function's rate of change is zero or undefined, indicating potential critical points.
To find these points, set the first derivative equal to zero:

• For our function: \( -2\sin(2x) - 1 = 0 \)
• Simplifying, \( \sin(2x) = -\frac{1}{2} \)

Solving for \( x \) involves considering angles where \( \sin \theta = -\frac{1}{2} \). In this case, \( x = \frac{7\pi}{12} + k\pi \) and \( x = \frac{11\pi}{12} + k\pi \) are solutions when accounting for the periodicity of the trigonometric sine function.

Second Derivative Test

The Second Derivative Test provides insight into the concavity of a function around critical points, thus helping us determine whether these points are local minima, maxima, or neither. After computing the first derivative, we then take the second derivative to further analyze the function \( y = \cos(2x) - x \). The second derivative is: \( y'' = -4\cos(2x) \).
Once you have the second derivative, you can assess the nature of each critical point you found through the first derivative step:

• If \( y'' < 0 \), the function is concave down at this point, indicating a local maximum.
• If \( y'' > 0 \), the function is concave up, showing a local minimum.

For instance, at \( x = \frac{7\pi}{12} \), the second derivative is \( y''(\frac{7\pi}{12}) = -2 \), implying a local maximum. Meanwhile, at \( x = \frac{11\pi}{12} \), \( y''(\frac{11\pi}{12}) = 2 \), which indicates a local minimum. This test provides a visual and mathematical understanding of the function's behavior near these key points.

Trigonometric Functions in Calculus

Trigonometric functions such as sine and cosine often appear in calculus problems and are essential in finding critical points or changes in a graph's behavior. In our function \( y = \cos(2x) - x \), we used trigonometry to simplify the derivative expression and solve for \( x \). The periodic nature of trigonometric functions allows us to find an infinite number of solutions to equations such as \( \sin(2x) = -\frac{1}{2} \):

• They repeat values over regular intervals, specifically every \(2\pi\),
• This periodical repetition informs us that solutions are not just single values but a series of possible answers across the number line.

Mastering how to manipulate and analyze these functions with derivatives is crucial for students as they work on understanding how function graphs behave, where they reach maximum or minimum heights, and where those crucial turning points occur.