Problem 62 - University Calculus: Early Transcendentals

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

Step-by-Step Solution

Answer

The function has local maxima at \( x = -\pi \) and \( x = \pi \), and a local minimum at \( x = 0 \).

Step 1: Find the derivative of the function

First, we need to find the derivative of the function \( f(x) = -2\cos x - \cos^2 x \). Using the chain rule and the derivative of cosine, we get: \[ f'(x) = 2\sin x + 2\cos x\sin x = 2\sin x (1 + \cos x) \]

Step 2: Set the derivative to zero

To find the critical points, set \( f'(x) = 0 \) and solve for \( x \): \[ 2\sin x (1 + \cos x) = 0 \] This equation is satisfied when \( \sin x = 0 \) or \( \cos x = -1 \).

Step 3: Solve for critical points

For \( \sin x = 0 \), within the interval \([-\pi, \pi]\), \( x = -\pi, 0, \pi \). For \( \cos x = -1 \), within the same interval, \( x = \pi \). However, \( \pi \) was already found as a solution when \( \sin x = 0 \). So the critical points are \( x = -\pi, 0, \pi \).

Step 4: Evaluate the function at critical points

To identify local extrema, evaluate \( f(x) \) at the critical points: - \( f(-\pi) = -2(-1) - (-1)^2 = 1 \)- \( f(0) = -2(1) - (1)^2 = -3 \)- \( f(\pi) = -2(-1) - (-1)^2 = 1 \)

Step 5: Determine local extrema

By evaluating \( f(x) \) at the critical points, we find that - \( x=0 \) has a local minimum at \( f(0) = -3 \).- \( x=-\pi \) and \( x=\pi \) have local maxima at \( f(-\pi) = f(\pi) = 1 \).

Step 6: Graph the function and the derivative

Plot the function \( f(x) = -2\cos x - \cos^2 x \) over \([-\pi, \pi]\) and its derivative \( f'(x) = 2\sin x(1 + \cos x) \). Observe that:- Where \( f'(x) \) changes sign, \( f(x) \) has local extrema.- \( f(x) \) tends to increase where \( f'(x) > 0 \) and decrease where \( f'(x) < 0 \).

Step 7: Comment on the behavior of the function

When \( f'(x) = 0 \), \( f(x) \) has local minima and maxima. Between \( x = -\pi \) and \( x = 0 \), the derivative is negative, which means \( f(x) \) decreases, and between \( x = 0 \) and \( x = \pi \), the derivative is positive, indicating \( f(x) \) increases.

Key Concepts

Derivative of Trigonometric Function

To tackle the given function involving trigonometric elements, we need to find its derivative. Let's simplify the function:

• The function is given by: \[ f(x) = -2\cos x - \cos^2 x \]
• Here we have cosine and squared cosine terms making the function's relationship more complex.

The derivative process involves special rules for trigonometric functions:

• Using the fact that the derivative of \(\cos x\) is \(-\sin x\), we get the derivative of related terms.
• We apply the chain rule when differentiating the \(\cos^2 x\) part.
• Thus, applying these to each component, the derivative becomes: \[ f'(x) = 2\sin x + 2\cos x \sin x = 2\sin x (1 + \cos x) \]

This expression is key for our next steps in finding critical points and understanding the function's behavior.

Critical Points

Critical points are locations on a graph where the derivative is zero or undefined. For our function:

• We want to solve \( f'(x) = 0 \) to find these critical points.
• Our equation \( 2\sin x (1 + \cos x) = 0 \) gives potential solutions.
• The critical values of \( x \) are found where \( \sin x = 0 \) or \( \cos x = -1 \).
• Within the interval \([-\pi, \pi]\), these occur at \( x = -\pi, 0, \pi \).

These critical points tell us where potential local maxima or minima could occur depending on the second derivative or further analysis could be conducted.

Graph Analysis

Graphing the function alongside its derivative can give us insight into how the function behaves:

• The graph of \( f(x) \) and its derivative help reveal where the function reaches its peaks and valleys.
• At the critical points, the derivative \( f'(x) \) changes its sign, indicating a possible extremum.
• By evaluating the function at these critical points (\( -\pi, 0, \pi \)), we can verify that:
  • At \( x = -\pi \) and \( x = \pi \), we see local maxima where \( f(x) = 1 \).
  • At \( x = 0 \), we find a local minimum with \( f(x) = -3 \).

In general, understanding this behavior helps confirm that the derivative being zero corresponds to potential local maxima/minima.

Chain Rule

The chain rule is a vital tool in calculus used for finding the derivative of composite functions:

• It states that if a function \( y = g(f(x)) \), then the derivative \( y' \) is found as \( g'(f(x)) \cdot f'(x) \).
• For our function, \( \cos^2 x = (\cos x)^2 \), this requires applying the chain rule.
• We differentiate \( (\cos x)^2 \) by treating it as \( g(f(x)) = u^2 \) with \( u = \cos x \). Then:
  • The derivative of \( u^2 \) is \( 2u \cdot u' \).
  • Combining with \( u' = -\sin x \), we get \( 2\cos x (-\sin x) = -2\sin x \cos x \).
• This intricacy highlights the value of the chain rule in capturing the complexity of nested functions.

Understanding and applying the chain rule correctly allows for accurate derivative calculation, vital for identifying changes in the function's behavior.