Problem 14 - Thomas Calculus in SI Units

Question

Answer the following questions about the functions whose derivatives are given in Exercises \(1-14:\) a. What are the critical points of \(f ?\) b. On what open intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$f^{\prime}(x)=(\sin x+\cos x)(\sin x-\cos x), 0 \leq x \leq 2 \pi$$

Step-by-Step Solution

Answer

Critical points: \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\). Increasing on \([0, \frac{\pi}{4})\), \((\frac{5\pi}{4}, \frac{7\pi}{4})\). Local maxima at \(\frac{\pi}{4}, \frac{5\pi}{4}\); minima at \(\frac{3\pi}{4}, \frac{7\pi}{4}\).

Step 1: Factorize the Derivative

The given derivative is \(f'(x) = (\sin x + \cos x)(\sin x - \cos x)\). Expand this product using the identity \((a+b)(a-b) = a^2 - b^2\). Thus, \(f^{\prime}(x) = \sin^2 x - \cos^2 x\).

Step 2: Find the Critical Points

Critical points occur where \(f^{\prime}(x) = 0\) or where \(f^{\prime}(x)\) is undefined. Here, we solve \( \sin^2 x - \cos^2 x = 0\), which simplifies to \( \sin^2 x = \cos^2 x \). Solving \( \sin x = \pm \cos x \), we find the critical points to be \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 3: Determine Increasing/Decreasing Intervals

Test intervals around the critical points: \([0, \frac{\pi}{4}), (\frac{\pi}{4}, \frac{3\pi}{4}), (\frac{3\pi}{4}, \frac{5\pi}{4}),\) and \((\frac{5\pi}{4}, \frac{7\pi}{4}), (\frac{7\pi}{4}, 2\pi]\). For example, if we test \(x = \frac{\pi}{6}\) in \([0, \frac{\pi}{4})\), we find \(f'(\frac{\pi}{6}) > 0\), meaning \(f(x)\) is increasing. Similarly, test other intervals to find where \(f(x)\) is increasing (\([0, \frac{\pi}{4})\) and \((\frac{5\pi}{4}, \frac{7\pi}{4})\)) and where it is decreasing (\((\frac{\pi}{4}, \frac{3\pi}{4})\) and \((\frac{3\pi}{4}, \frac{5\pi}{4})\)).

Step 4: Identify Local Maxima and Minima

Using the First Derivative Test, evaluate the behavior of \(f(x)\) around the critical points. At \(x = \frac{\pi}{4}\), \(f'\) changes from positive to negative indicating a local maximum. At \(x = \frac{3\pi}{4}\), \(f'\) changes from negative to positive indicating a local minimum. Repeat for other critical points: \(x = \frac{5\pi}{4}\) gives a local maximum and \(x = \frac{7\pi}{4}\) gives a local minimum.

Key Concepts

Critical Points

Critical points in a function are crucial because they might be where the function reaches its peaks and valleys. To find critical points, we need to set the derivative of the function, denoted by \( f'(x) \), equal to zero and solve for \( x \). This is where the slope of the tangent is flat, hinting at a possible peak or trough.

In our example, we took the derivative \( f'(x) = \sin^2 x - \cos^2 x \), and solved \( \sin^2 x = \cos^2 x \). Solving this equation led us to identify the critical points \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \). Critical points can also occur where the derivative is undefined, but for our derivative, this was not the case. Finding these points helps us understand not just where but why changes in behavior occur in a function.

Increasing and Decreasing Intervals

An increasing interval of a function is where its derivative is greater than zero, which indicates that the function is rising as we move from left to right on the graph. Conversely, in a decreasing interval, the derivative is less than zero, showing the function is falling.

To determine these intervals for the function, we examine the sign of \( f'(x) \) between the critical points:

• For \([0, \frac{\pi}{4})\), testing a point such as \( x = \frac{\pi}{6} \) gives \( f'\left(\frac{\pi}{6}\right) > 0\), so the function is increasing.
• Between \( (\frac{\pi}{4}, \frac{3\pi}{4}) \), test an \( x \) such as \( x = \frac{\pi}{2} \) which gives \( f'\left(\frac{\pi}{2}\right) < 0\), indicating the function is decreasing.

Following this approach for each interval around our critical points allows us to map out where the function grows or declines across its domain.

Local Maxima and Minima

Local maxima and minima are points where a function reaches a peak or a valley, respectively. To find these, we examine the sign changes of \( f'(x) \) around the critical points.

Using the First Derivative Test, a point \( x = c \) is a local maximum if \( f'(x) \) changes from positive to negative at \( c \). A local minimum occurs when \( f'(x) \) changes from negative to positive.

• For example, at \( x = \frac{\pi}{4} \), \( f'(x) \) goes from positive, indicating an increasing function, to negative, signaling a decrease. Thus, \( x = \frac{\pi}{4} \) is a local maximum.
• At \( x = \frac{3\pi}{4} \), the derivative changes from negative to positive, so it becomes a local minimum.

This analysis provides a deeper understanding of how the function behaves and helps in sketching the graph with accuracy.