Problem 79 - Thomas Calculus

Question

In Exercises \(75-80,\) you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$ f(x)=\sqrt{x}+\cos x, \quad[0,2 \pi] $$

Step-by-Step Solution

Answer

The absolute extrema are found at the endpoints and critical points using f(x) values.

Step 1: Plot the function

Use a CAS (Computer Algebra System) to plot the function \( f(x) = \sqrt{x} + \cos x \) over the interval \([0, 2\pi]\). This will help you visualize the function's behavior and identify any potential extrema.

Step 2: Find interior points where \( f'(x) = 0 \)

First, find the derivative of the function: \( f'(x) = \frac{1}{2\sqrt{x}} - \sin x \). Set it to zero: \( \frac{1}{2\sqrt{x}} - \sin x = 0 \). Solve this equation using a numerical solver to find the interior points within the interval \([0, 2\pi]\).

Step 3: Plot and Find Points where \( f'(x) \) does not exist

Identify where the expression \( f'(x) = \frac{1}{2\sqrt{x}} - \sin x \) is undefined. The derivative is undefined when \( x = 0 \) because \( \sqrt{x} \) in the denominator becomes zero, making the derivative not exist at \( x = 0 \).

Step 4: Evaluate the function at critical points and endpoints

Compute \( f(x) \) at the critical points found where \( f'(x) = 0 \), at points where the derivative does not exist, and at the endpoints \( x = 0 \) and \( x = 2\pi \). For example, compute \( f(0) \), \( f(2\pi) \), and other values as needed.

Step 5: Determine absolute extreme values

Compare the function values obtained in Step 4. Identify the maximum and minimum values, which represents the absolute extrema of \( f(x) = \sqrt{x} + \cos x \) on the interval \([0, 2\pi]\). State these extrema and the points at which the occur.

Key Concepts

Critical Points

Critical points are locations on a graph where the derivative is zero or does not exist. These points are essential in calculus as they often indicate places where the function could have a local maximum, minimum, or saddle point. To find these points, we first take the derivative of a function. For the function given, \( f(x) = \sqrt{x} + \cos x \), the derivative is \( f'(x) = \frac{1}{2\sqrt{x}} - \sin x \).
Critical points occur where this derivative equals zero. Solving \( \frac{1}{2\sqrt{x}} - \sin x = 0 \) will uncover such points within the interval \([0, 2\pi]\).
Additionally, critical points also occur where the derivative does not exist. For \( f'(x) \), the term \( \frac{1}{2\sqrt{x}} \) is undefined at \( x = 0 \), creating another critical point. Understanding these points is crucial for determining where the function reaches its highest and lowest values.

Absolute Extrema

Absolute extrema refer to the highest and lowest values a function attains on a given interval. They consist of the absolute maximum and minimum values.
To find these, we examine several things:

• Critical points inside the interval where the derivative is zero or undefined.
• The endpoints of the interval, since extrema can occur at the start or end of an interval.

Once these points are identified, we compute the function's value at each. Comparing these values will reveal the absolute extrema.
In our function over the interval \([0, 2\pi]\), we need the function’s values at all critical points and endpoints to determine the absolute minimum and maximum values. These extreme values provide important insights, such as the peak point and lowest dip on a graph.

Derivatives

Derivatives are a key concept in calculus, representing the rate of change of a function. They give us insight into the behavior of a function, such as when it is increasing or decreasing, or when the slope of the function equals zero.
For \( f(x) = \sqrt{x} + \cos x \), the derivative \( f'(x) = \frac{1}{2\sqrt{x}} - \sin x \) shows how \( f(x) \) changes with respect to \( x \).
A positive derivative implies the function is increasing, while a negative derivative indicates it is decreasing. Zero indicates potential extrema, while undefined derivatives often reveal boundary behaviors even more compelling.
In our context, derivatives help us find critical points, which are essential for identifying any absolute extrema on a specified interval.

Closed Interval

A closed interval is a range of values that includes its endpoints. It is denoted as \([a, b]\), meaning it spans from \( a \) to \( b \) and includes both \( a \) and \( b \).
For problems involving absolute extrema, the closed interval plays a vital role, as it confines where we look for extreme values.
In the given exercise, the closed interval \([0, 2\pi]\) specifies where we need to look for critical points and where we evaluate the endpoints. This ensures that we don’t miss potential boundary extrema, maintaining the completeness of our solution. By analyzing this interval, one ensures all absolute maximum and minimum values are accounted for within the specified range.