Problem 37 - Calculus Early Transcendentals

Question

Analyze the trigonometric function f over the specified interval, stating where f is increasing, decreasing, concave up, and concave down, and stating the x-coordinates of all inflection points. Confirm that your results are consistent with the graph of f generated with a graphing utility. \(f(x)=(\sin x+\cos x)^{2} ;[-\pi, \pi]\)

Step-by-Step Solution

Answer

The function is increasing on \( (\frac{-\pi}{4}, \frac{3\pi}{4}) \), decreasing elsewhere; it is concave up on \( (-\pi,0) \cup (\pi,0) \), concave down on \( (0,\pi) \). Inflection points occur at \( x = 0 \), \( x = \pi \).

Step 1: Understand the function

We are given the function \( f(x) = (\sin x + \cos x)^2 \) over the interval \( [-\pi, \pi] \). Our task is to determine where the function is increasing, decreasing, concave up, concave down, and identify any inflection points.

Step 2: Find the first derivative

To determine where \( f(x) \) is increasing or decreasing, calculate its derivative: \( f'(x) = 2(\sin x + \cos x)(\cos x - \sin x) \). This results from applying the chain rule to the original function.

Step 3: Analyze the first derivative

Set \( f'(x) = 0 \) to find critical points where the function changes direction. Solve \( 2(\sin x + \cos x)(\cos x - \sin x) = 0 \), leading to the systems \( \sin x + \cos x = 0 \) or \( \cos x - \sin x = 0 \). These can be simplified to find specific angles within the interval \([-\pi, \pi]\).

Step 4: Determine intervals of increase and decrease

From the solutions of the first derivative, check intervals determined by these critical points to see where \( f'(x) \) is positive (increasing) or negative (decreasing). Test values between critical points in \( f'(x) \) to establish behavior.

Step 5: Find the second derivative

To determine concavity, find the second derivative: \( f''(x) \). Use the product rule on \( f'(x) \), resulting in \( f''(x) = 2[(\cos x - \sin x)^2 + (\sin x + \cos x)(-\sin x - \cos x)] \). Simplify this expression for ease of use.

Step 6: Analyze the second derivative

Set \( f''(x) = 0 \) to find potential inflection points. Solve for \( x \) to locate where concavity might change. These are points where \( f''(x) \) changes sign.

Step 7: Determine intervals of concavity

Test intervals based on critical points from \( f''(x) = 0 \) to establish where \( f(x) \) is concave up (\( f''(x) > 0 \)) or concave down (\( f''(x) < 0 \)). Confirm any changes in concavity which verify inflection points.

Key Concepts

Derivative Analysis

Derivative analysis helps us understand how a function behaves regarding increase, decrease, and its rate of change. The derivative of a function, denoted as \( f'(x) \), represents the slope of the tangent to the function at any point. Here, for the function \( f(x) = (\sin x + \cos x)^2 \), we need to determine where it is increasing or decreasing.

To find \( f'(x) \), we apply the chain rule:

• First, differentiate the outer function, \( (\sin x + \cos x)^2 \), to get \( 2(\sin x + \cos x) \).
• Then, differentiate the inner function, \( \sin x + \cos x \), giving \( \cos x - \sin x \).

By multiplying these results, we find:\[ f'(x) = 2(\sin x + \cos x)(\cos x - \sin x)\]Understanding this derivative helps us pinpoint where the original function increases (when \( f'(x) > 0 \)) or decreases (when \( f'(x) < 0 \)). By solving \( f'(x) = 0 \), we also locate critical points which are the stepping stones for the next analyses.

Critical Points

Critical points occur where the derivative is zero or undefined because the slope of the tangent turns horizontal, indicating potential local maxima, minima, or inflection points. For \( f(x) = (\sin x + \cos x)^2 \), solve:\[ f'(x) = 2(\sin x + \cos x)(\cos x - \sin x) = 0\]To find these critical points:

• Set \( \sin x + \cos x = 0 \): solving gives the angles \( x = -\frac{3\pi}{4}, \frac{{\pi}{4} \).
• Set \( \cos x - \sin x = 0 \): solving gives \( x = -\frac{\pi}{4}, \frac{\pi}{4} \).

Each critical point needs examination within the interval \([-\pi, \pi]\) to determine how they contribute to the behavior of the function. Evaluating these can reveal where the function’s direction changes and highlights key transition points.

Concavity and Inflection Points

Concavity tells us how the curve bends or reshapes itself. If a curve is concave up, it looks like a smile, and if it's concave down, it looks like a frown. Inflection points are where the function changes its concavity.
We use the second derivative, \( f''(x) \), to analyze concavity:\[ f''(x) = 2[(\cos x - \sin x)^2 + (\sin x + \cos x)(-\sin x - \cos x)]\]Solving \( f''(x) = 0 \) gives us the x-values where concavity may switch and potential inflection points. Test values in these intervals:

• If \( f''(x) > 0 \), the function is concave up.
• If \( f''(x) < 0 \), the function is concave down.

Where \( f''(x) \) changes sign, locate inflection points, providing peaks of switches in curvature. These points are significant as they give insight into the function's overall structure.

Interval Analysis

Interval analysis integrates the results from derivative analysis and critical points to provide comprehensive insights into where the function rises, falls, and bends. It involves scanning between critical points and inflection points:

• Check interval behavior by testing specific values and determining if the first derivative is positive or negative. Positive suggests an increasing section; negative a decrease.
• For concavity, observe interval sections determined by the second derivative. Verify if \( f''(x) \) results in concave up or down in those zones.

This step-by-step approach uncovers patterns illustrating where the function spikes, flattens, or shifts on its curve. Verifying these results against graphical tools can ensure mathematical work's accuracy—a vital step as visual confirmations reinforce theoretical deductions and signal real-world applications.