Problem 29 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ f(x)=\tan x / 2 ;(-\pi / 2, \pi / 6) $$

Step-by-Step Solution

Answer

Minimum: None (approaches \(-\infty\) at \(x \to-\pi/2^+\)), Maximum: \(\frac{1}{\sqrt{3}}\) at \(x=\pi/6\).

Step 1: Find the derivative of the function

To find the critical points where the function may have extreme values, we first need to find the derivative of the function \( f(x) = \tan(x/2) \). By using the chain rule and the fact that the derivative of \( \tan u \) is \( \sec^2 u \), we find:\[ f'(x) = \frac{1}{2} \sec^2(x/2) \]

Step 2: Determine where the derivative is undefined or zero

Extreme values occur where the derivative is zero or undefined. The derivative \( f'(x) = \frac{1}{2} \sec^2(x/2) \) is never zero because the secant squared function doesn't equal zero. However, it is undefined where \( \sec(x/2) \rightarrow \infty \), which occurs when \( x/2 = \frac{\pi}{2} + k\pi \) or equivalently when \( x = \pi + 2k\pi \). We need to find such points in the interval.

Step 3: Evaluate where the derivative is undefined within the interval

In the interval \( (-\pi/2, \pi/6) \), we convert \( \pi + 2k\pi \) to find points where \( x = \pm \pi/2 \), but \( x = 0, \pi, \ldots \) are outside the given interval. Therefore, we only need to check endpoints and any undefined points calculated earlier.

Step 4: Evaluate the function at the endpoints of the interval

Since the function \( f(x)=\tan(x/2) \) is continuous on \( (-\pi/2, \pi/6) \) except at the endpoints, we evaluate the limits as \( x \) approaches the boundaries. As \( x \to -\pi/2^+ \), \( \tan(x/2) \to -\infty \) and as \( x \to \pi/6^- \), \( \tan(x/2) \to \frac{1}{\sqrt{3}} \).

Step 5: Determine the extreme values

Since the derivative does not provide any zero points and the only issues occur at boundaries, we can conclude: The minimum is \(-\infty\) as \( x \to -\pi/2^+ \) and maximum value is \( \frac{1}{\sqrt{3}} \) occurring at \( x = \pi/6 \).

Key Concepts

Extreme Values

When discussing extreme values in calculus, we often refer to the maximum and minimum values a function can attain on a given interval. These values provide insights into the behavior and limits of the function.

To find extreme values, we investigate the points where the function has the potential to reach its peaks or troughs. Generally, these points are located where the derivative of the function equals zero or is undefined. At these points, the slope of the tangent to the function is horizontal (0) or non-existent.

For the function \( f(x) = \tan(x/2) \) within the interval \((-\pi/2, \pi/6)\), we find that the derivative \( f'(x) \) does not become zero, but it is undefined at certain points. This leads us to focus on the interval's boundaries and any undefined points to find extreme values.

Evaluating the endpoints helps us determine that the function approaches \(-\infty\) as \( x \) approaches \(-\pi/2^+\) and achieves a maximum value of \( \frac{1}{\sqrt{3}} \) as \( x \) approaches \( \pi/6^-\). Thus, the extreme values are reflected as a minimum of \(-\infty\) and a maximum of \( \frac{1}{\sqrt{3}} \) in this context.

Derivative

Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to its variable. They are essential for understanding how functions behave and finding extreme values.

To find the derivative of \( f(x) = \tan(x/2) \), we utilize the chain rule. This rule allows differentiation of composite functions by finding the derivative of the outer function and multiplying it by the derivative of the inner function. Given that the derivative of \( \tan u \) is \( \sec^2 u \), we apply this knowledge to obtain

• \( f'(x) = \frac{1}{2} \sec^2(x/2) \).

To determine possible extreme values, we inspect where \( f'(x) \) might be zero (horizontal tangents) or undefined. In this example, the derivative is undefined, which means the function has vertical tangents or is discontinuous at certain points. Specifically, the undefined spots occur where \( \sec(x/2) \rightarrow \infty \), indicating possible extremes or asymptotic behavior.

Understanding derivatives thus gives us critical information about the nature of the function, allowing us to assess where changes happen most acutely or where the function might present potential extreme values.

Trigonometric Functions

Trigonometric functions like tangent play a significant role in calculus. They are periodic and have distinct behaviors that make them unique in their analysis.

For the function \( f(x) = \tan(x/2) \), we see an example of these properties. Tangent, like the sine and cosine functions, deals with angles and is periodic. However, unlike sine and cosine, tangent has vertical asymptotes, points where the function goes off to infinity, recurring every \( \pi \) units, making it particularly interesting when discussing extreme values.

When analyzing \( \tan(x/2) \), we note

• Its period is expanded compared to \( \tan x \)..
• Vertical asymptotes occur at \( x = \pi + 2k\pi \) for integer \( k \), leading to places where the function is undefined within the interval.

The function's behavior incorporates both the periodic nature of trigonometric functions and the asymptotic within every period, prompting careful consideration when evaluating extreme values.

By examining trigonometric functions, we not only gain insight into their intrinsic properties but also apply these insights to specific cases like finding extreme values within defined intervals.