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Chapter 6: Problem 78

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

Expert verified
#Short Answer# To find the critical points of \(f(x) = \frac{\cosh x}{x}\), first find the derivative, \(f'(x) = \frac{\sinh x \cdot x - \cosh x}{x^2}\). Then, find where \(f'(x) = 0\) or is undefined and show that these critical points satisfy the equation \(x = \operatorname{coth} x\). Finally, approximate the critical points using a root finder (like Newton's method, the bisection method, or a root-finding tool) on the function \(g(x) = x - \operatorname{coth} x\).

Step by step solution

01

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

To find the critical points of \(f(x)\), we first need to find its derivative, \(f'(x)\). Using the quotient rule, we have: $$f'(x) = \frac{(\cosh x)' \cdot x - \cosh x \cdot (x)'}{x^2}$$ Recall that the derivative of \(\cosh x\) is \(\sinh x\), and the derivative of \(x\) is \(1\). Therefore, $$f'(x) = \frac{\sinh x \cdot x - \cosh x \cdot 1}{x^2}$$
02

Find where \(f'(x)\) is equal to zero or undefined

To find the critical points, we need to find where \(f'(x) = 0\) or is undefined. \(f'(x)\) is undefined when the denominator is zero, i.e., when \(x = 0\). However, the original function \(f(x)\) is also undefined at \(x=0\), so we disregard that value. Now we need to solve: $$\frac{\sinh x \cdot x - \cosh x}{x^2} = 0$$
03

Show that the critical points satisfy \(x = \operatorname{coth} x\)

To show the critical points satisfy \(x = \operatorname{coth} x\), we need to manipulate the equation \(f'(x) = 0\). We can rewrite \(\sinh x \cdot x - \cosh x = 0\) as: $$x \sinh x = \cosh x$$ Divide both sides by \(\cosh x\): $$x = \frac{\cosh x}{\sinh x}$$ Recall that \(\operatorname{coth} x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x}\), so we have shown: $$x = \operatorname{coth} x$$
04

Approximate the critical points using a root finder

To approximate the critical points, use a root finder (such as Newton's method, the bisection method, or a root-finding tool) on the function: $$g(x) = x - \operatorname{coth} x$$ Determine starting values close to where the critical points might be. For this example, let's choose \(x = 1\) for our starting value, and proceed using the selected root-finding method until we obtain a desired level of accuracy. The root will be an approximation of the critical point of the function \(f(x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbolic Functions
Hyperbolic functions, much like their trigonometric counterparts, are essential in various fields of mathematics and science.
They can often describe the shapes of natural phenomena and appear in the solutions to certain differential equations.
The most common hyperbolic functions are the hyperbolic sine and cosine, denoted by \( \sinh x \) and \( \cosh x \), respectively.
  • Hyperbolic Sine (\( \sinh x \)): Defined as \( \sinh x = \frac{e^x - e^{-x}}{2} \). It has properties similar to the sine function, but it grows exponentially.
  • Hyperbolic Cosine (\( \cosh x \)): Defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \). This function always returns positive values and resembles the shape of the usual cosine curve when plotted.
  • Hyperbolic Cotangent (\( \operatorname{coth} x \)): Given by \( \operatorname{coth} x = \frac{\cosh x}{\sinh x} \). This function relates \( x \) to the hyperbolic functions, serving as a bridge for solving equations as seen in the problem.
In calculus, these functions are critical because they allow us to explore problems with more complexity, offering insights into behaviors that are not possible with simple trigonometric functions.
Quotient Rule
The quotient rule is a vital tool in calculus used for differentiating functions that are expressed as ratios.
Given two differentiable functions, \( u(x) \) and \( v(x) \), the derivative of their quotient, \( \frac{u(x)}{v(x)} \), is calculated using the formula:
\[ \left( \frac{u}{v} \right)' = \frac{u' \cdot v - u \cdot v'}{v^2} \]
This rule helps us differentiate complex expressions involving division without error.
Here鈥檚 how it works step-by-step:
  • First, differentiate the numerator function \( u(x) \) to get \( u'(x) \).
  • Then, differentiate the denominator function \( v(x) \) to get \( v'(x) \).
  • Multiply \( u'(x) \) by \( v(x) \), this will be the first term in the formula.
  • Multiply \( u(x) \) by \( v'(x) \), this will be subtracted from the first term.
  • Finally, divide the whole expression by \( v^2(x) \).
This technique was applied to find the critical points of the function \( f(x) = \frac{\cosh x}{x} \), where the derivative \( f'(x) \) was computed to assist in the analysis of critical points.
Root Finding Methods
Root finding methods are an essential part of numerical analysis, aiding in solving equations where analytical solutions are difficult or impossible to find.
These methods estimate the value of \( x \) for which a function \( g(x) = 0 \). Here are some common root finding methods that you can use:
  • Newton's Method: This iterative approach uses tangents to approximate the roots. Starting from an initial guess, \( x_0 \), the next approximation is calculated as \( x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} \).
  • Bisection Method: This method narrows the interval where the root lies by repeatedly bisecting it and selecting the subinterval where the sign of the function changes.
  • Secant Method: Related to Newton's method but does not require the derivative. It uses two initial guesses and approximates the root using secant lines.
In the given exercise, a root finding tool was applied to the function \( g(x) = x - \operatorname{coth} x \) to approximate the critical points of \( f(x) \).
This process helps in determining where the critical points occur by finding approximate solutions.

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