Question

Use Newton's method to find all local extreme values of \(f(x)=x \operatorname{sech} x\).

Short answer

Answer: 1. Find the first derivative of the function. 2. Set the first derivative equal to zero and apply Newton's method. 3. Find the second derivative of the function. 4. Implement Newton's method using the second derivative to find critical points. 5. Determine if the critical points are maximum or minimum by analyzing the second derivative.

Step-by-step solution

Find the derivative of the function f(x)

To find the critical points, we first need to find the derivative of the function \(f(x) = x \operatorname{sech}(x)\). Using the product and chain rule, we have: \( f'(x) = \operatorname{sech}(x) - x \operatorname{sech}(x) \operatorname{tanh}(x)\)

Set the derivative equal to zero and apply Newton's method

To find the critical points, we need to set the first derivative of the function equal to zero: \( 0= \operatorname{sech}(x) - x \operatorname{sech}(x) \operatorname{tanh}(x)\) Now, we can use Newton's method to find the zeros of the first derivative function. Newton's method is an iterative method that starts with an initial guess, \(x_0\), and generates the next guess using the formula: \(x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)}\) Here, \(x_{n+1}\) is the updated value, \(x_n\) was the previous guess, and \(f''(x_n)\) is the second derivative of \(f\). To use Newton's method, we first need to find the second derivative of the function.

Find the second derivative of the function f(x)

To find the second derivative, we differentiate the first derivative: \(f''(x) = -2\operatorname{sech}(x)\operatorname{tanh}(x)+x[(2\operatorname{tanh}^2(x)-1)\operatorname{sech}(x)-\operatorname{sech}(x)\tanh^2(x)]\) Now we can use Newton's method to find the zeros of the first derivative function.

Implement Newton's method

Choose an initial guess value of \(x_0\) to start the iteration process. We will need to perform a few iterations to find an accurate value for the critical point (it might be necessary to start with different initial guess for multiple critical points). Let us start with an initial guess of \(x_0 = 0.5\): Repeat the iteration process using the formula: \(x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)}\) Keep iterating until \(x_n\) converges to an accurate critical point value.

Determine if the critical point is maximum or minimum

Once the critical points have been identified, determine if they represent local maximums or minimums by analyzing the second derivative of the function. If \(f''(x) > 0\) at a critical point, then the function has a local minimum at that point. If \(f''(x) < 0\) at a critical point, then the function has a local maximum at that point. Use the critical point value obtained in the previous step and evaluate the second derivative, \(f''(x)\), at this point to determine whether it is a local maximum or minimum. By following these steps, you can find all local extreme values of the function \(f(x) = x \operatorname{sech}(x)\) using Newton's method.

Key concepts

Local Extreme Values

Understanding local extreme values is essential when analyzing the behavior of a function. A local extreme value can either be a maximum or a minimum within a particular interval. These points are significant because they represent where a function changes direction or exhibits a peak or valley.

When analyzing a function like \(f(x) = x \operatorname{sech}(x)\), identifying local extreme values helps us know where the function's slope becomes zero, changing from increasing to decreasing, or vice versa. This information is crucial in various mathematical and practical applications, such as optimization problems.

• Local Maximum: The highest point in the immediate neighborhood of the function.
• Local Minimum: The lowest point in the immediate neighborhood of the function.

By understanding how to find these points, particularly through Newton's method, you gain powerful insights into the structure of the function.

Derivative

The derivative of a function is a fundamental tool in calculus that provides insights into the function's behavior. It tells us the rate at which a function is changing at any given point. To find critical points, we need to determine where this rate of change reaches zero, which corresponds to where the function has a potential local extreme value.

For the function \(f(x) = x \operatorname{sech}(x)\), we use the product and chain rules to find the first derivative:

• Product Rule: Used when differentiating products of two functions.
• Chain Rule: Used when differentiating compositions of functions.

The derivative is given by:\[ f'(x) = \operatorname{sech}(x) - x \operatorname{sech}(x) \operatorname{tanh}(x).\]By solving \(f'(x) = 0\), we locate potential critical points for further analysis.

Critical Points

Critical points are where the first derivative of a function is zero or undefined. They are important because they might indicate local maxima or minima. We identify these points by setting the first derivative equal to zero and solving the resulting equation.

Upon obtaining the derivative \( f'(x) = \operatorname{sech}(x) - x \operatorname{sech}(x) \operatorname{tanh}(x) \), solving \( f'(x) = 0 \) gives potential critical points.

• These aren't always extreme values; further testing is needed.
• Critical points are crucial markers for applying Newton's method.

With these points in hand, we use Newton’s method to refine our guesses for the locations where extreme values occur.

Second Derivative

The second derivative of a function helps us determine the nature of the critical points by indicating concavity of the function at those points. This derivative tells us about the curvature direction:

• If \(f''(x) > 0\), the function is concave up, indicating a local minimum.
• If \(f''(x) < 0\), the function is concave down, indicating a local maximum.

For \(f(x) = x \operatorname{sech}(x)\), the second derivative is calculated as:\[f''(x) = -2\operatorname{sech}(x)\operatorname{tanh}(x) + x[(2\operatorname{tanh}^2(x)-1)\operatorname{sech}(x)-\operatorname{sech}(x)\tanh^2(x)].\]Evaluating this at critical points determined by the first derivative assists in confirming whether these points are maxima or minima through concavity analysis. Understanding these concepts together aids in effectively using Newton's method for practical problem-solving.