Problem 42 - Calculus Early Transcendentals

Question

Use a graphing utility to estimate the absolute maximum and minimum values of \(f \), if any, on the stated interval, and then use calculus methods to find the exact values. \(f(x)=\cos (\sin x) ;[0, \pi]\)

Step-by-Step Solution

Answer

Absolute maximum is 1 at \(x = 0, \pi\); absolute minimum is \(\cos 1\) at \(x = \frac{\pi}{2}\).

Step 1: Understand the Problem

We are given a function \( f(x) = \cos(\sin x) \) and need to determine its absolute maximum and minimum on the interval \([0, \pi]\). First, use a graphing utility to estimate these values visually.

Step 2: Use a Graphing Utility

Using a graphing utility, plot the function \( f(x) = \cos(\sin x) \) on the interval \([0, \pi]\). Visually inspect the graph to identify any peaks (maximums) and troughs (minimums) within the given interval.

Step 3: Identify Critical Points Analytically

To find critical points, differentiate \( f(x) = \cos(\sin x) \). Using the chain rule, we find \( f'(x) = -\sin(\sin x) \cdot \cos x \). Set \( f'(x) = 0 \) and solve for \( x \) to identify the critical points.

Step 4: Solve for Critical Points

Solving \(-\sin(\sin x) \cdot \cos x = 0\) gives us two equations: \( \sin(\sin x) = 0 \) and \( \cos x = 0 \). \( \sin(\sin x) = 0 \) implies \( \sin x = 0 \), so \( x = 0 \) in the interval, and \( \cos x = 0 \) implies \( x = \frac{\pi}{2} \). Thus, critical points are \( x = 0 \) and \( x = \frac{\pi}{2} \).

Step 5: Evaluate the Function at Critical Points and Endpoints

Evaluate \( f(x) = \cos(\sin x) \) at the critical points and the endpoints: \( x = 0, \frac{\pi}{2}, \pi \). We have \( f(0) = \cos 0 = 1 \), \( f(\frac{\pi}{2}) = \cos 1 \approx 0.5403 \), and \( f(\pi) = \cos 0 = 1 \).

Step 6: Determine Absolute Maximum and Minimum

By comparing the values, \( f(x) = 1 \) at \( x = 0 \) and \( x = \pi \), which are maximums, and \( f(x) \approx 0.5403 \) at \( x = \frac{\pi}{2} \) is the minimum value. Conclude that the absolute maximum is 1 and the absolute minimum is \( \cos 1 \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that is used to differentiate composite functions. When you have a function nestled within another function, the chain rule helps you find the derivative.
For the function in the exercise, \( f(x) = \cos(\sin x) \), the chain rule is employed because \( \sin x \) is nested inside \( \cos \). To differentiate such a function, you differentiate the outer function and multiply it by the derivative of the inner function.

• The derivative of \( \cos(u) \) is \( -\sin(u) \).
• And the derivative of \( \sin x \) is \( \cos x \).

Thus, applying the chain rule, the derivative \( f'(x) \) is calculated as \( -\sin(\sin x) \cdot \cos x \). This helps us find critical points by setting the derivative equal to zero.

Graphing Utility

Using a graphing utility can greatly assist in visualizing the behavior of a function over a given interval. Graphing calculators or software, such as Desmos or GeoGebra, allow us to explore functions and their critical points.
By plotting \( f(x) = \cos(\sin x) \) across the interval \([0, \pi]\), you can manually observe where the function reaches its peaks and valleys. This visual inspection provides an initial estimate of where the maximum and minimum values occur, though a precise calculus-based approach is also needed.
A graphing utility can complement calculus techniques by confirming the locations of critical points we've calculated analytically. However, for exact values, analytical methods are required.

Critical Points

Finding critical points involves setting the derivative of a function to zero. Critical points are where the function's slope is zero, indicating potential maximum or minimum values.
In our function, \( f'(x) = -\sin(\sin x) \cdot \cos x \), we set this equal to zero and solve:

• \( -\sin(\sin x) = 0 \Rightarrow \sin x = 0 \), which implies \( x = 0 \) within the interval.
• \( \cos x = 0 \Rightarrow x = \frac{\pi}{2} \).

These critical points, along with the endpoints of the interval, are where you evaluate the function to determine any maximums or minimums. This process verifies what we've seen visually and allows for exact calculation of function values at these important points.

Absolute Maximum and Minimum

Absolute maximum and minimum refers to the highest and lowest values that a function can take within a specified interval. These are important to identify as they describe the range of function values in practical scenarios.
After pinpointing the critical points and plotting them against the endpoints, you evaluate the function \( f(x) = \cos(\sin x) \) at \( x = 0 \), \( x = \frac{\pi}{2} \), and \( x = \pi \):

• At \( x = 0 \), \( f(0) = \cos 0 = 1 \) (maximum).
• At \( x = \frac{\pi}{2} \), \( f(\frac{\pi}{2}) = \cos 1 \approx 0.5403 \) (minimum).
• At \( x = \pi \), \( f(\pi) = \cos 0 = 1 \) (maximum).

Here, the absolute maximum value of the function on the interval is 1, and the absolute minimum is approximately 0.5403. Understanding these values helps when interpreting the behavior of solutions over an interval.