Problem 58 - Thomas Calculus

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 \(f^{\prime}. \) $$ f(x)=-2 \cos x-\cos ^{2} x, \quad-\pi \leq x \leq \pi $$

Step-by-Step Solution

Answer

The local minimum is at \( x = 0 \) and local maxima are at \( x = \pm \pi \). Graphs show \( f'(x) \) changes sign indicating these points are extrema.

Step 1: Find the derivative

To find the local extrema, we first need to find the derivative of the function. Given the function \[ f(x) = -2 \cos x - \cos^2 x \] Use the chain rule and the power rule to find the derivative: \[ f'(x) = (-2 \cos x)' - (\cos^2 x)' \] \[ = 2 \sin x + 2\cos x \sin x \] This simplifies to: \[ f'(x) = 2 \sin x ( 1 + \cos x ) \]

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. Here, the derivative is defined everywhere on the interval, so set it equal to zero: \[ 2 \sin x (1 + \cos x) = 0 \] This gives us two equations to solve: 1. \( \sin x = 0 \) 2. \( 1 + \cos x = 0 \) From \( \sin x = 0 \), we get \( x = 0, \pm \pi \). From \( \cos x = -1 \), we get \( x = \pm \pi \). Thus, the critical points are \( x = 0, \pm \pi \).

Step 3: Evaluate at critical points and endpoints

Evaluate the function at critical points and endpoints to find the local extrema: - For \( x = 0 \), \[ f(0) = -2\cos(0) - \cos^2(0) = -2(1) - (1)^2 = -3 \] - For \( x = \pi \), \[ f(\pi) = -2\cos(\pi) - \cos^2(\pi) = 2(1) - (1)^2 = 1 \] Evaluate at \( x = -\pi \) similarly as it's the same calculation as \( x = \pi \): - For \( x = -\pi \), \[ f(-\pi) = 1 \] Thus, the function has a local minimum at \( x = 0 \), and local maxima at \( x = \pi \) and \( x = -\pi \).

Step 4: Graph the function and derivative

Sketch the graphs of \( f(x) = -2 \cos x - \cos^2 x \) and its derivative \( f'(x) = 2\sin x (1 + \cos x) \). Observe: - At \( x = 0 \), since \( f'(x) = 0 \) and changes sign across zero (from negative to positive), \( f \) has a local minimum. - At \( x = \pi \) and \( x = -\pi \), \( f'(x) = 0 \) and does not change sign across zero, indicating possible local maxima.

Step 5: Comment on the function's behavior

The value of \( f'(x) \) indicates the slope of the tangent line at any point on \( f(x) \). - Where \( f'(x) > 0 \), \( f(x) \) is increasing. - Where \( f'(x) < 0 \), \( f(x) \) is decreasing. In the interval \( -\pi \leq x \leq \pi \), \( f(x) \) decreases to \( x = 0 \), then increases to \( x = \pi \), consistent with local minimum and maxima points identified.

Key Concepts

Derivative

Finding the derivative of a function is essential when looking for local extrema. The derivative of a function, denoted as \( f'(x) \), provides us with the slope of the tangent line at any point on \( f(x) \). To calculate the derivative, we can use rules like the chain rule and power rule. In this exercise, we found the derivative of \( f(x) = -2 \cos x - \cos^2 x \) to be \( f'(x) = 2 \sin x (1 + \cos x) \). This derivative tells us where the function is increasing or decreasing:

• When \( f'(x) > 0 \), the function is increasing, meaning the slope of the tangent is positive.
• When \( f'(x) < 0 \), the function is decreasing, indicating the slope of the tangent is negative.
• Local extrema occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined, although in this case, \( f'(x) \) is defined everywhere on the interval.

Finding the derivative correctly is the first step to solving for local extrema.

Critical Points

Critical points are key when analyzing the behavior of \( f(x) \) for potential local minima and maxima. Critical points occur at values of \( x \) where the derivative \( f'(x) = 0 \) or where \( f'(x) \) is undefined. In this exercise, the derivative \( f'(x) = 2 \sin x (1 + \cos x) \) was set to zero to find critical points.

• We solved \( 2 \sin x (1 + \cos x) = 0 \) to find \( x = 0, \pm \pi \) as the critical points.
• These critical points were tested in the function to find the local minima and maxima.

The fact that we can solve for critical points so explicitly demonstrates the straightforward nature of our derivative, which is continuous and defined over the entire interval \(-\pi \leq x \leq \pi\).

Graphing Functions

Visualizing a function and its derivative on a graph can provide better insight into the function's behavior. By graphing \( f(x) = -2 \cos x - \cos^2 x \) and its derivative \( f'(x) = 2 \sin x (1 + \cos x) \), you can:

• See where the slope of \( f(x) \) is changing, indicating local minima and maxima.
• Observe how \( f(x) \) behaves in relation to the signs of \( f'(x) \).
• Identify points where the derivative intersects the x-axis, corresponding to critical points.

In this particular problem, the graph shows that \( f(x) \) has a local minimum at \( x = 0 \) and local maxima at \( x = \pi \) and \( x = -\pi \). Graphing enhances understanding by connecting algebraic analysis to visual information.

Sin and Cos Functions

Both sine and cosine functions are fundamental trigonometric functions that play a vital role in this exercise. Understanding their properties can make finding derivatives and solving equations simpler.

• The function \( \cos x \) oscillates between -1 and 1, with key points at multiples of \( \pi \).
• Similarly, \( \sin x \) also oscillates between -1 and 1 but has zeros at multiples of \( \pi \).
• These periodic properties are vital when solving equations like \( \sin x = 0 \) or \( 1 + \cos x = 0 \), which are crucial for finding critical points.

Knowing these properties simplifies the fact that \( \sin x = 0 \) gives us \( x = 0, \pm \pi \) and \( \cos x = -1 \) provides us with \( x = \pm \pi \). These important trigonometric rules support finding solutions to complex calculus problems.