Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ f(x)=\sin ^{2} 2 x \text { on }[0,2] $$

Step-by-Step Solution

Answer

The global maximum is 1 and the global minimum is 0 on \([0, 2]\).

Step 1: Determine Critical Points within the Interval

The first step is to find the critical points of the function within the interval \([0, 2]\). We start by differentiating the function. The function is given by \(f(x)=\sin^2(2x)\). Let's apply the chain rule to differentiate this: \(f'(x) = 2\sin(2x)\cdot\cos(2x)\cdot2 = 4\sin(2x)\cos(2x)\). Using the double angle identity, we rewrite this as \(f'(x) = 2\sin(4x)\). Set the derivative equal to zero to find the critical points: \[2\sin(4x) = 0\] Solving \(\sin(4x)=0\) gives \(4x = n\pi\), where \(n\) is an integer. Thus, \(x=n\pi/4\). List these points in \([0, 2]\): \(x=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, 1, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\).

Step 2: Evaluate the Function at Critical Points and Endpoints

Next, evaluate \(f(x)\) at each critical point and at the endpoints of the interval, \([0, 2]\). Recall that \(f(x)=\sin^2(2x)\). Calculate: - \(f(0)=\sin^2(0)=0\) - \(f\left(\frac{\pi}{4}\right)=\sin^2\left(\frac{\pi}{2}\right)=1\) - \(f\left(\frac{\pi}{2}\right)=\sin^2(\pi)=0\) - \(f\left(\frac{3\pi}{4}\right)=\sin^2\left(\frac{3\pi}{2}\right)=1\) - \(f(1)=\sin^2(2)\approx0.826\)- \(f\left(\frac{5\pi}{4}\right)=\sin^2\left(\frac{5\pi}{2}\right)=1\)- \(f\left(\frac{3\pi}{2}\right)=\sin^2(3\pi)=0\)- \(f\left(\frac{7\pi}{4}\right)=\sin^2\left(\frac{7\pi}{2}\right)=1\)- \(f(2)=\sin^2(4)\approx0.573\).

Step 3: Identify the Global Maximum and Minimum

From the calculations in Step 2, we need to identify the highest and lowest values of \(f(x)\) on \([0, 2]\). The maximum value is \(1\), occurring at \(x=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\). The minimum value is \(0\), occurring at \(x=0, \frac{\pi}{2}, \frac{3\pi}{2}\).

Key Concepts

Critical Points

Critical points are essentially the values of \(x\) in a given function where the derivative either is zero or doesn't exist. These points are crucial, as they often help us identify where the function changes direction, peaking at a local maximum or dipping to a local minimum.
When finding critical points, especially in the context of global extrema, we focus on the given interval. In our case, for the function \(f(x) = \sin^2(2x)\) over the interval \([0, 2]\), we started by differentiating \(f(x)\). Using the chain rule as shown in our solution, we derived the function and set the derivative equal to zero, which led us to the equation \(2\sin(4x) = 0\).

This implies \(\sin(4x) = 0\), and the critical points occur at values of \(x\) where \(4x = n\pi\), giving \(x = n\pi/4\) (with \(n\) as an integer). Evaluating this formula within the interval \([0, 2]\), we determined the possible critical values: \(x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, 1, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\).

• Use calculus tools like derivative to find critical points.
• Critical points help determine potential max/min points.
• A derivative of zero implies a stationary point.

Interval Evaluation

Interval evaluation involves analyzing a function within a specific range of its domain to find global extremes, which may be located at critical points or at the interval’s endpoints.
After determining the critical points for \(f(x) = \sin^2(2x)\) within the interval \([0, 2]\), the next step was to evaluate the function at these critical points and the interval boundaries, specifically at \(x = 0\) and \(x = 2\).

This step is essential because the maximum and minimum values of \(f(x)\) on \([0, 2]\) could occur at any of these points. Evaluating \(f(x)\) at these values involves substituting each point back into the original function. We calculated:

• \(f(0) = 0\)
• \(f\left(\frac{\pi}{4}\right) = 1\)
• \(f\left(\frac{\pi}{2}\right) = 0\)
• \(f\left(\frac{3\pi}{4}\right) = 1\)
• \(f(1) \approx 0.826\)
• \(f\left(\frac{5\pi}{4}\right) = 1\)
• \(f\left(\frac{3\pi}{2}\right) = 0\)
• \(f\left(\frac{7\pi}{4}\right) = 1\)
• \(f(2) \approx 0.573\)

This systematic evaluation ensures we don't miss any potential extrema points, thereby allowing us to confidently identify the global maxima and minima of the function over the specified interval.

Sine Function

The sine function, denoted as \(\sin(x)\), is a periodic function that oscillates between -1 and 1. This makes it useful in modeling waves and cyclical patterns.
In our scenario with \(f(x) = \sin^2(2x)\), we are looking at a transformed version of the sine function. Here, the argument of the sine function is doubled, \(2x\), affecting its period and the square of the sine causing the output range to be between 0 and 1.

Every complete cycle of \(\sin(x)\) returns to its starting point, making the sine function predictable and repetitive. When squaring the sine function, as is the case in \(\sin^2(2x)\), the negative values of sine are rendered positive, causing the graph of \(f(x)\) to be always non-negative.

• The frequency doubling in \(\sin(2x)\) results in a halved period compared to \(\sin(x)\).
• Squaring the function ensures all values are zero or positive.
• Recognize the impact of transformations on the sine's amplitude and frequency.

Understanding these transformations is critical in effectively analyzing the function \(f(x) = \sin^2(2x)\) and identifying potential global extrema within a specified interval.