Problem 60 - Thomas Calculus in SI Units

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)=\sec ^{2} x-2 \tan x, \quad \frac{-\pi}{2}

Step-by-Step Solution

Answer

Local minimum at \( x = \frac{\pi}{4} \); \( f \) decreases before and increases after this point.

Step 1: Find the Derivative

To find the local extrema of the function \( f(x) = \sec^2 x - 2 \tan x \), we start by finding its derivative, \( f'(x) \). Using the derivatives \( \frac{d}{dx}[\sec x] = \sec x \tan x \) and \( \frac{d}{dx}[\tan x] = \sec^2 x \):\[ f'(x) = 2\sec x \sec x \tan x - 2\sec^2 x = 2\sec^2 x \tan x - 2\sec^2 x = 2 \sec^2 x (\tan x - 1)\]

Step 2: Solve for Critical Points

Set the derivative \( f'(x) = 0 \) to find the critical points:\[2 \sec^2 x (\tan x - 1) = 0\]Dividing both sides by \( 2 \sec^2 x \), we need to solve \( \tan x - 1 = 0 \), which gives:\[\tan x = 1\]Thus, \( x = \frac{\pi}{4} \), considering the given interval \( \frac{-\pi}{2} < x < \frac{\pi}{2} \).

Step 3: Analyze the Interval

We need to check the behavior of \( f'(x) \) around \( x = \frac{\pi}{4} \):- For \( x < \frac{\pi}{4} \), choose \( x = 0 \): \( \tan 0 = 0 \), \( 2 \sec^2 0 (\tan 0 - 1) = -2 \); hence, \( f'(x) < 0 \).- For \( x > \frac{\pi}{4} \), choose \( x = \frac{\pi}{3} \): \( \tan \frac{\pi}{3} = \sqrt{3} \), \( 2 \sec^2 \frac{\pi}{3} (\tan \frac{\pi}{3} - 1) > 0 \); hence, \( f'(x) > 0 \).This indicates \( f(x) \) decreases on \( x < \frac{\pi}{4} \) and increases on \( x > \frac{\pi}{4} \), so \( x = \frac{\pi}{4} \) is a local minimum.

Step 4: Graph and Interpret Behavior

Graph \( f(x) = \sec^2 x - 2 \tan x \) and its derivative \( f'(x) = 2 \sec^2 x (\tan x - 1) \). The graph shows that \( f(x) \) reaches a minimum at \( x = \frac{\pi}{4} \) and increases as \( x \) moves away from \( \frac{\pi}{4} \). The derivative \( f'(x) \) transitions from negative to positive at this point, consistent with a local minimum behavior.

Key Concepts

Derivative of Trigonometric Functions

To find the local extrema of a function like \( f(x) = \sec^2 x - 2 \tan x \), the first step is to compute its derivative, \( f'(x) \). In this case, we utilize known derivatives of trigonometric functions:

• The derivative of \( \sec x \) is \( \sec x \tan x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).

Combining these, the derivative of \( f(x) \) becomes:\[ f'(x) = 2\sec^2 x \tan x - 2\sec^2 x \].
This simplifies further to:
\[ f'(x) = 2 \sec^2 x (\tan x - 1) \].
This derivative tells us how the function changes at each point for \( x \) within the interval \( \frac{-\pi}{2} < x < \frac{\pi}{2} \). Understanding these derivatives is crucial, as they guide us in determining where the function might have local minima or maxima.

Calculating derivatives of trigonometric functions forms the foundation for more complex behavior analysis, ensuring precision when finding critical points.

Critical Points

Critical points occur where the function's derivative, \( f'(x) \), equals zero or is undefined. These points are potential locations for local extrema, making them essential in our interval analysis.

For our function \( f(x) \), solving the equation:\[ 2 \sec^2 x (\tan x - 1) = 0 \]
leads us to isolate factors where \( \tan x - 1 = 0 \).
This simplifies to \( \tan x = 1 \), and within our interval \( \frac{-\pi}{2} < x < \frac{\pi}{2} \), the solution is \( x = \frac{\pi}{4} \).

Finding critical points is an essential step in discerning where the function could potentially have turning points or points of inflection. Once these points are identified, further analysis reveals more about the function's local behavior.

Behavior of Functions

Understanding the behavior of a function around critical points is key to determining local extrema. This involves assessing the values of \( f'(x) \) at various positions relative to the critical point.

For \( x = \frac{\pi}{4} \):

• Consider \( x < \frac{\pi}{4} \): setting \( x = 0 \), gives \( \tan 0 = 0 \), making \( f'(x) = -2 \), which is negative.
• For \( x > \frac{\pi}{4} \): setting \( x = \frac{\pi}{3} \), \( \tan \frac{\pi}{3} = \sqrt{3} \), yields a positive \( f'(x) \).

This analysis shows \( f(x) \) decreases before \( x = \frac{\pi}{4} \) and increases after, indicating a local minimum.

Observing this transition is crucial because the behavior of \( f'(x) \) provides insight into whether a function reaches peak points or valleys at these critical values.

Interval Analysis

Interval analysis involves interpreting function behavior across different sections of its domain. For function \( f(x) = \sec^2 x - 2 \tan x \), the interval \( \frac{-\pi}{2} < x < \frac{\pi}{2} \) needs careful examination.

By dividing the interval around critical points, like \( x = \frac{\pi}{4} \), we check neighboring intervals using sample values to determine the sign of \( f'(x) \).

• If \( f'(x) < 0 \), the function is decreasing.
• If \( f'(x) > 0 \), the function is increasing.

In our analysis:

• For \( x < \frac{\pi}{4} \), \( f(x) \) was found decreasing.
• For \( x > \frac{\pi}{4} \), \( f(x) \) was found increasing.

This interval-specific behavior ultimately confirms \( x = \frac{\pi}{4} \) as a local minimum. Interval analysis consolidates findings from derivatives and critical points, offering a full picture of the function's behavior.