Problem 38 - 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. \(f(x)=\sin ^{2} 2 x ;[0, \pi]\)

Step-by-Step Solution

Answer

Increasing: (0, π/4), (π/2, 3π/4); Decreasing: (π/4, π/2), (3π/4, π); Concave up: (π/8, 3π/8), (5π/8, 7π/8); Concave down: (0, π/8), (3π/8, 5π/8), (7π/8, π); Inflection points: π/8, 3π/8, 5π/8, 7π/8.

Step 1: Find the derivative

To analyze the function, we need to find its derivative. Let's use the chain rule and derivative of \(\sin(x)\) to find \(f'(x)=\frac{d}{dx}(\sin^2(2x))\).\Let \(u=\sin(2x)\), then \(f(x)=u^2\) and \(f'(x)=2u \cdot u'\).Now, find \(u'=\cos(2x)\cdot 2\).Thus, \(f'(x)=2\sin(2x) \cdot 2 \cos(2x) = 4\sin(2x)\cos(2x)\).Simplifying using the double angle identity, \(f'(x)=2\sin(4x)\).

Step 2: Identify critical points

Set the first derivative equal to zero to find critical points: \(2\sin(4x)=0\).Thus, \(\sin(4x) = 0\).The solutions for \(\sin\theta = 0\) occur at \( heta = k\pi\), where \(k\) is an integer.Therefore, \(4x = k\pi\) and \(x = \frac{k\pi}{4}\), within \(x = [0, \pi]\).This gives: \(x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\).

Step 3: Determine intervals of increase/decrease

Use test points in the intervals defined by critical points to determine if \(f(x)\) is increasing or decreasing.Test intervals: \([0, \frac{\pi}{4}], [\frac{\pi}{4}, \frac{\pi}{2}], [\frac{\pi}{2}, \frac{3\pi}{4}], [\frac{3\pi}{4}, \pi]\).For example, in \(x \in (0, \frac{\pi}{4})\), pick \(x = \frac{\pi}{8}\). Since \(\sin(\frac{\pi}{2}) > 0\), \(f'(x) > 0\) implies \(f(x)\) is increasing.Repeat for each interval.

Step 4: Find the second derivative

Find the second derivative to determine concavity.Start with \(f'(x) = 2\sin(4x)\).Using the chain rule, \(f''(x)=2 \cdot 4\cos(4x)=8\cos(4x)\).

Step 5: Determine concavity

Set the second derivative \(f''(x) = 8\cos(4x)\) equal to zero to find points of inflection: \(8\cos(4x)=0\).Solve for \(\cos(4x) = 0\).The solutions for \(\cos\theta = 0\) occur at \(\theta = \frac{\pi}{2} + k\pi\).Thus, \(4x = \frac{\pi}{2} + k\pi\) gives \(x= \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}\).Determine concavity by testing intervals between points of inflection.

Key Concepts

Derivative Calculation

Calculating the derivative of a function is a fundamental operation in calculus. Here, we need to find the derivative of the given trigonometric function, \(f(x) = \sin^2(2x)\). To do this, we apply the chain rule. The chain rule helps us differentiate composite functions. In our case:

• Let \( u = \sin(2x) \). Then \( f(x) = u^2 \).
• The derivative of \( u^2 \) with respect to \( x \) is \( 2u \cdot u' \).
• Next, we find \( u' = \cos(2x) \cdot 2 \), because the derivative of \( \sin(2x) \) is \( \cos(2x) \cdot 2 \).
• Combine these results to get \( f'(x) = 4\sin(2x)\cos(2x) \).

Using the double angle identity, \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), simplifies \( f'(x) \) to \( 2\sin(4x) \). Derivatives allow us to analyze the behavior of functions, such as finding where they increase or decrease.

Critical Points

Critical points occur where the first derivative of a function is zero or undefined. These points are vital in understanding the function's behavior as they can help identify where a function changes from increasing to decreasing or vice versa. For the function \( f(x) = \sin^2(2x) \), its derivative is \( f'(x) = 2\sin(4x) \).
To find the critical points, set

• \( 2\sin(4x) = 0 \).
• Thus, \( \sin(4x) = 0 \).
• The solutions occur at \( 4x = k\pi \), where \( k \) is an integer.
• Solving for \( x \), we get \( x = \frac{k\pi}{4} \).

Within the interval \([0, \pi]\), the critical points are \( x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi \). These points divide the interval into sections we can evaluate to understand where the function increases or decreases.

Concavity Analysis

Concavity tells us how the shape of the graph behaves whether it is curving upwards or downwards. To explore the concavity, we calculate the second derivative of the function. We begin with the first derivative \( f'(x) = 2\sin(4x) \) and apply the chain rule again to find the second derivative:

• The derivative of \( \sin(4x) \) is \( \cos(4x) \cdot 4 \).
• This makes \( f''(x) = 2 \cdot 4\cos(4x) = 8\cos(4x) \).

With \( f''(x) \), you determine the concavity:

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

Testing around critical x-values helps confirm if each interval of the function is correctly concave up or down.

Inflection Points

Inflection points mark where a graph changes its concavity, switching between concave up and concave down. To find these points for our function, set the second derivative equal to zero: \( f''(x) = 8\cos(4x) = 0 \).

• Thus, \( \cos(4x) = 0 \).
• The solutions occur at \( 4x = \frac{\pi}{2} + k\pi \).
• This gives us \( x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \) within the interval \([0, \pi]\).

Testing the intervals between these x-values helps determine the concavity's changes. These inflection points tell us where the graph of the function shifts its curve direction, a critical insight for understanding overall graph behavior.