Question

Critical points a. Show that the critical points of \(f(x)=\frac{\cosh x}{x}\) satisfy \(x=\operatorname{coth} x\) b. Use a root finder to approximate the critical points of \(f\)

Short answer

#Answer# The critical points of the function \(f(x) = \frac{\cosh x}{x}\) satisfy the equation \(x = \coth x\). One approximate critical point is \(x \approx 2.026\).

Step-by-step solution

Understanding critical points

Critical points are points on the graph of a function where its derivative is either zero or undefined. These points help us understand the behavior of the function, as they represent its relative maxima, minima, or points of inflection.

Find the derivative of \(f(x)\)

To find the critical points, we first need to find the derivative of \(f(x)\): $$f(x) = \frac{\cosh x}{x}$$ Apply the quotient rule: $$f'(x) = \frac{x \cdot \frac{d}{dx}(\cosh x) - \cosh x \cdot \frac{d}{dx}(x)}{x^2}$$ Now the derivatives of \(\cosh x\) and \(x\) are \(\sinh x\) and \(1\) respectively, so substitute them: $$f'(x) = \frac{x \cdot \sinh x - \cosh x}{x^2}$$

Show critical points satisfy \(x = \coth x\)

To find the critical points, we need to set the derivative \(f'(x)\) equal to zero: $$0 = \frac{x \cdot \sinh x - \cosh x}{x^2}$$ Multiply both sides by \(x^2\) to eliminate the denominator: $$0 = x \cdot \sinh x - \cosh x$$ To rewrite it as \(x = \coth x\), we have to isolate \(x\) on one side: $$x = \frac{\cosh x}{\sinh x}$$ Notice that \(\coth x = \frac{\cosh x}{\sinh x}\), therefore the critical points of \(f(x)\) satisfy: $$x = \coth x$$

Use a root finder to approximate critical points

To approximate the critical points of \(f\), we can use a root finder like the Newton-Raphson method or bisection method for the equation \(x = \coth x\). You can implement the method in a programming language or use software like MATLAB, Python, or Mathematica to find approximations of the critical points. For example, using a numerical solver, we may find that one critical point is approximately \(x \approx 2.026\)

Key concepts

Quotient Rule

The quotient rule allows us to differentiate functions that are expressed as a quotient of two other functions, say \( u(x) \) and \( v(x) \). If you have a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative is given by:\[f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\]Here's a breakdown of the formula:

• \( v(x) \) is the function in the denominator.
• \( u'(x) \) and \( v'(x) \) are the derivatives of the numerator and denominator functions, respectively.

In the given problem, we applied the quotient rule to find the derivative of \( f(x) = \frac{\cosh x}{x} \). So, \( \cosh x \) is \( u(x) \) and \( x \) is \( v(x) \). With derivatives \( \sinh x \) and \( 1 \), the rule helps provide the expression \( f'(x) = \frac{x \cdot \sinh x - \cosh x}{x^2} \).Remember, the quotient rule helps reveal critical points where the derivative simplifies to zero or is undefined, thus revealing key behavior changes in the function.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but for a hyperbola. They include \( \sinh x \), \( \cosh x \), and other related functions like \( \tanh x \) and \( \coth x \). Their definitions are rooted in exponential functions:

• \( \cosh x = \frac{e^x + e^{-x}}{2} \)
• \( \sinh x = \frac{e^x - e^{-x}}{2} \)

These functions are crucial in modelling scenarios where growth occurs consistently with hyperbolic attributes, such as in certain physics and engineering contexts.In our problem, we used these functions to simplify the derivative calculation and express the critical point condition \( x = \coth x \). Notice that \( \coth x = \frac{\cosh x}{\sinh x} \), recognizing this can help in precising transformations and solutions in differential problems. As with trigonometric functions, understanding their derivatives and behaviors is vital for effectively utilizing hyperbolic identities.

Numerical Methods

Numerical methods involve algorithms to solve mathematical problems that cannot be easily solved analytically. These methods approximate solutions using iterative procedures. They are especially useful for finding roots of equations, optimizing functions, or handling complex integrations.To approximate critical points like \( x = \coth x \), numerical techniques become essential. Methods such as the Newton-Raphson or bisection method repeatedly adjust values in pursuit of convergence to the actual solution.For example, the Newton-Raphson method uses iterations with:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]This formula shifts the current guess \( x_n \) towards a zero of the function \( f(x) \), making it a powerful tool for root-finding problems.Numerical methods require an initial estimate and knowledge of the function's behavior, thus underpinning entire technological and scientific problem-solving landscapes. In computations, ensure you deploy these methods when encountering complex or unsolvable algebraic equations.