Problem 45 - Calculus Early Transcendentals

Question

If \(f\) is a periodic function, then the locations of all absolute extrema on the interval \((-\infty,+\infty)\) can be obtained by finding the locations of the absolute extrema for one period and using the periodicity to locate the rest. Use this idea in these exercises to find the absolute maximum and minimum values of the function, and state the \(x\) -values at which they occur. \(f(x)=2 \cos x+\cos 2 x\)

Step-by-Step Solution

Answer

The absolute maximum is 3 at \(x = 2n\pi\) and the minimum is -1 at \(x = (2n+1)\pi\), for integer \(n\).

Step 1: Identify the period of the function

The function is given by \(f(x) = 2 \cos x + \cos 2x\). Since \(\cos x\) and \(\cos 2x\) are periodic, we find the period by considering the least common multiple of their periods. \(\cos x\) has a period of \(2\pi\) and \(\cos 2x\) has a period of \(\pi\), so the least common multiple is \(2\pi\). Therefore, \(f(x)\) is periodic with period \(2\pi\).

Step 2: Find critical points in one period

To find critical points, differentiate \(f(x)\): \(f'(x) = -2\sin x - 2\sin 2x\). Set \(f'(x) = 0\) to find critical points: \(-2\sin x - 2(2\sin x \cos x) = 0\). Simplifying gives \(-2\sin x (1 + 2\cos x) = 0\). This implies \(\sin x = 0\) or \(\cos x = -\frac{1}{2}\). Solve for \(x\) in the interval \([0, 2\pi]\).

Step 3: Evaluate at critical points and endpoints

The critical points within \([0, 2\pi]\) are \(x = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}\). Calculate \(f(x)\) at these points and at the endpoints of the period, \(x = 0, 2\pi\): - At \(x = 0\), \(f(x) = 2 + 1 = 3\). - At \(x = \pi\), \(f(x) = -2 + 1 = -1\). - At \(x = \frac{2\pi}{3}\), \(f(x) = 2(-\frac{1}{2}) + (-\frac{1}{2}) = -\frac{3}{2}\). - At \(x = \frac{4\pi}{3}\), \(f(x) = 2(-\frac{1}{2}) + (-\frac{1}{2}) = -\frac{3}{2}\). - At \(x = 2\pi\), \(f(x) = 3\).

Step 4: State the absolute maximum and minimum

From the evaluations, the maximum value in one period \([0, 2\pi]\) is \(3\) at \(x = 0, 2\pi\). The minimum value is \(-1\) at \(x = \pi\). Due to periodicity, this pattern repeats, so the maximum values occur at \(x = 2n\pi\) and the minimum values at \(x = (2n+1)\pi\) for integers \(n\).

Key Concepts

Critical Points

Critical points of a function are where the derivative either equals zero or does not exist. These points are important for determining where a function's graph changes direction, thus helping us identify potential locations for extrema.
For trigonometric functions, this involves differentiating the function and setting it to zero to solve for critical points. In the case of our function, the derivatives involve trigonometric identities that simplify our computation.
Once you set the derivative of our trigonometric function, which in this case is \(-2\sin x - 2\sin 2x\) equal to zero, you solve for \(x\) within the fundamental period (\([0, 2\pi]\)). Here, the simplification yields critical points at \(x = 0, \pi, \frac{2\pi}{3},\) and \(\frac{4\pi}{3}\). These are the candidates for where maximum and minimum values might be located within one period.

• Critical points occur where a function's derivative is zero.
• Knowing trigonometric identities can simplify the process.

Absolute Extrema

Absolute extrema refer to the highest or lowest values of a function over a specified interval or its entire domain. Finding these involves looking at both critical points and endpoints. For periodic functions, you only need to examine values within one period, then extrapolate to the full domain using the function's periodicity.
In the given function \(f(x) = 2 \cos x + \cos 2x\), after finding the critical points, you then evaluate the function's value at these points within one period (\([0, 2\pi]\)) to see which ones produce the highest and lowest values. For instance, \(f(x)\) was evaluated at \(x = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi\).
The results showed an absolute maximum value of 3 occurring at \(x = 0, 2\pi\), and an absolute minimum value of -1 occurring at \(x = \pi\) within one period.

• Absolute maxima and minima provide the overall highest and lowest function values.
• These extrema must account for critical points and periodicity.

Periodicity of Trigonometric Functions

The periodicity of a trigonometric function refers to its repetitive nature over certain intervals, meaning it repeats its pattern at fixed intervals known as periods.
This property is crucial for functions like \(f(x)=2 \cos x + \cos 2x\), where multiple terms each have a distinct period. To work with such functions, you typically calculate the least common multiple (LCM) of the individual periods. Since \(\cos x\) has a period of \(2\pi\) and \(\cos 2x\) has a period of \(\pi\), the combined period \(f(x)\) becomes \(2\pi\), the LCM of the two periods.
Understanding this allows you to evaluate the function only within a single period, confidently knowing the pattern will repeat indefinitely over its domain. This means you only need to identify critical points and extrema once, then apply that information over all periods in the domain.

• Trigonometric functions repeat their values over defined intervals.
• Using the least common multiple helps find a unified period.