Problem 120 - Thomas Calculus in SI Units

Question

Where does the periodic function \(f(x)=2 e^{\sin (x / 2)}\) take on its extreme values and what are these values?

Step-by-Step Solution

Answer

\( f(x) \) reaches maximum \( 2e \) at \( x = \pi + 4k\pi \) and minimum \( \frac{2}{e} \) at \( x = 3\pi + 4k\pi \).

Step 1: Understand the Function

The given function is \( f(x) = 2 e^{\sin(x/2)} \). It is a periodic function due to the sine component, \( \sin(x/2) \), which means it repeats its pattern over intervals. The maximum and minimum values of \( \sin(x/2) \) will determine the extreme values of \( f(x) \).

Step 2: Determine the Critical Points

To find the extreme values, we need to evaluate \( f(x) \) at the maximum and minimum points of \( \sin(x/2) \). The sine function, \( \sin(x/2) \), has a maximum value of 1 and a minimum value of -1 because the range of the sine function is always between -1 and 1.

Step 3: Find the Maximum Value

When \( \sin(x/2) = 1 \), \( f(x) = 2 e^1 = 2e \). To achieve \( \sin(x/2) = 1 \), we consider \( x/2 = \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer. Thus, the maximum value, \( 2e \), is taken at points like \( x = \pi + 4k\pi \).

Step 4: Find the Minimum Value

When \( \sin(x/2) = -1 \), \( f(x) = 2 e^{-1} = \frac{2}{e} \). To achieve \( \sin(x/2) = -1 \), we have \( x/2 = \frac{3\pi}{2} + 2k\pi \). Therefore, the minimum value, \( \frac{2}{e} \), occurs at points like \( x = 3\pi + 4k\pi \).

Step 5: Conclusion

The function \( f(x) = 2 e^{\sin(x/2)} \) takes on its maximum value of \( 2e \) when \( \sin(x/2) = 1 \), and its minimum value of \( \frac{2}{e} \) when \( \sin(x/2) = -1 \). These values occur at periodic intervals based on \( k \).

Key Concepts

Periodic Functions

Periodic functions are functions that repeat their values at regular intervals. They are a foundational concept in mathematics and widely used in physics and engineering. The pattern or shape of such a function's graph continues over the same interval repeatedly. This interval is called the period.
For example:

• The sine and cosine functions, with a period of \(2\pi\), are classic examples of periodic functions.
• In the context of the exercise, the function \(f(x) = 2 e^{\sin(x/2)}\) includes the periodic element \(\sin(x/2)\), which impacts the periodic behavior of the entire function.

Understanding the periodicity helps in finding extreme points over one period; these points repeat identically over subsequent intervals. Thus, determining extreme values becomes much more manageable.

Sine Function

The sine function is a smooth, wave-like function that oscillates between -1 and 1. As a result, it models repetitive phenomena like sound waves and light waves very well. Its maximum and minimum values are achieved at specific angles. This sinusoidal behavior can be expressed by:

• For sine \((\sin\theta)\), it reaches a maximum of 1 at angles like \(\frac{\pi}{2}, \frac{5\pi}{2}, ...\) and a minimum of -1 at \(\frac{3\pi}{2}, \frac{7\pi}{2}, ...\).
• In the exercise, the function's sine component \(\sin(x/2)\) modifies this pattern by dividing the angle \(x\) by 2, effectively stretching the wave horizontally. However, the values of \(\sin(x/2)\) remain within the same range of -1 to 1.

Evaluating the sine function’s maximum and minimum helps in determining the points where the original function \(f(x)\) has its extreme values. Understanding sine's behavior is crucial for analyzing changes and trends in periodic functions.

Critical Points

Critical points in a function are where the function could have extreme values, such as maximums or minimums. Detecting these points involves finding where the derivative of the function equals zero or is undefined, as these are potential locations for local maxima or minima.
For the exercise function \(f(x) = 2 e^{\sin(x/2)}\):

• The critical points occur when \(\sin(x/2) = 1\) or \(\sin(x/2) = -1\), as these are the locations where the sine component reaches its extremes.
• To find where this occurs, solve the equations \(\frac{x}{2} = \frac{\pi}{2} + 2k\pi\) and \(\frac{x}{2} = \frac{3\pi}{2} + 2k\pi\), yielding solutions for \(x\) as \(x = \pi + 4k\pi\) for the maximum and \(x = 3\pi + 4k\pi\) for the minimum, respectively.

Recognizing and calculating critical points allows for a comprehensive understanding of the function’s behavior and is crucial for identifying its extreme values.