Problem 38 - Thomas Calculus

Question

In Exercises \(37-40 :\) a. Find the local extrema of each function on the given interval, and say where they are assumed. 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

Local maxima at \( x = \pm \pi \), local minimum at \( x = 0 \).

Step 1: Find the Derivative

To find the local extrema, we first need to find the derivative of the function. Let's find \( f'(x) \) for \( f(x) = -2 \cos x - \cos^2 x \).The derivative is computed as follows:\[ f'(x) = \frac{d}{dx}(-2 \cos x - \cos^2 x) \]Using the chain rule and the fact that the derivative of \( \cos x \) is \( -\sin x \), we get:\[ f'(x) = 2 \sin x + 2 \cos x \sin x \, \Rightarrow \, 2 \sin x (1 + \cos x) \].

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative \( f'(x) = 2 \sin x (1 + \cos x) \) equal to zero and solve for \( x \).\[ 2 \sin x (1 + \cos x) = 0 \]This equation gives us two cases to consider: 1. \( \sin x = 0 \)2. \( 1 + \cos x = 0 \).

Step 3: Solve Case 1

For the first case \( \sin x = 0 \), we solve for \( x \): The solutions are \( x = 0, \pm \pi \) within the interval \([-\pi, \pi]\).

Step 4: Solve Case 2

For the second case \( 1 + \cos x = 0 \), solve for \( x \):\[ \cos x = -1 \]This yields the solution \( x = \pi \). However, since \( \pi \) was already found in Case 1, no new critical points emerge from this calculation.

Step 5: Evaluate at Critical Points

Evaluate \( f(x) = -2 \cos x - \cos^2 x \) at the critical points found: \( x = -\pi, 0, \pi \).- At \( x = -\pi \) and \( x = \pi \), \( f(x) = 0 \).- At \( x = 0 \), \( f(x) = -3 \).

Step 6: Determine Local Extrema

By comparing the values, \( f(x) = 0 \) at \( x = \pm \pi \) which are the local maxima, and \( f(x) = -3 \) at \( x = 0 \) which is the local minimum in the interval \([-\pi, \pi]\).

Step 7: Graph the Function and Derivative

Using graphing technology or software, plot \( f(x) = -2 \cos x - \cos^2 x \) and its derivative \( f'(x) = 2 \sin x (1 + \cos x) \) over the interval \([-\pi, \pi]\).Notice where \( f'(x) = 0 \), \( f(x) \) has local extrema. Where \( f'(x) > 0 \), \( f(x) \) is increasing, and where \( f'(x) < 0 \), \( f(x) \) is decreasing.

Step 8: Analyze the Behavior

The derivative \( f'(x) \) changes sign at the critical points indicating local extrema:- At \( x = 0 \), \( f'(x) \) changes from positive to negative, so \( f(x) \) has a local minimum.- At \( x = \pm \pi \), \( f'(x) \) changes from negative to positive, indicating local maxima.

Key Concepts

Derivative

The derivative of a function is a central concept in calculus that helps us understand how the function behaves. It's like a tool that tells us the rate at which the function's value is changing at any specific point. To find the derivative of a given function, we apply certain rules of differentiation, including the power rule, the product rule, and the chain rule, which we'll talk about later. In our problem, the function we dealt with was
\( f(x) = -2 \cos x - \cos^2 x \).
By differentiating, we got
\( f'(x) = 2 \sin x (1 + \cos x) \).
This expression tells us how the function's slope changes across different points on its graph. Finding the derivative gives us a critical step toward locating those special points known as the function's critical points.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These are the points where the function might have local maxima, minima, or might change its direction. In the exercise, to find the critical points, we set the derivative
\( f'(x) = 0 \),
which gives us an equation to solve. In our situation, this resulted in
\[ 2 \sin x (1 + \cos x) = 0 \].
Solving this, the points \( x = 0, \pm\pi \) were found within the interval
\([-\pi, \pi]\).
These points are essential as they tell us where the behavior of the function changes. By checking the value of our original function at these points, we can determine if they are offering local minima or maxima.

Interval

An interval is a range of values, often defined numerically, over which we are interested to examine the behavior of a function. In calculus, it’s common to find local extremes of functions within a specific interval. For our exercise, the given interval was
\([-\pi, \pi]\).
This means we're only considering critical points and changes in function behavior within that range.
Checking against the interval helps in ensuring that any critical points calculated are indeed within the scope of analysis, providing relevant maxima or minima that respect the function's domain under scrutiny.

Chain Rule

The chain rule is a powerful method in calculus used to differentiate composite functions. It essentially helps us break down the process when dealing with functions within functions. Think of it as calculating the derivative of the 'inner' function and the 'outer' function, then multiplying them together. In our exercise, the chain rule was applied to differentiate
\(-\cos^2 x\).
This involves recognizing that we have a chain where the outer function is squaring and the inner function is
\(-\cos x\).
The chain rule allowed us to efficiently find
our derivative, transforming the complexity of composition into manageable steps, leading to our critical derivative:
\(2 \sin x (1 + \cos x)\).
Taking derivatives of composite functions without this rule would be cumbersome, underscoring its vital role in calculus.